Tcl, 91 bytes

proc D n {while \$n {set n [expr ~$n&((1<<[string le [format %b $n]])-1)]
incr i}
incr i 0}

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proc D {n i\ 0} {while \$n {set n [expr ~$n&((1<<[string le [format %b $n]])-1)]
incr i}
set i}

Go, 172 bytes

import(."fmt";."strings")
func f(n uint)(i int){for;n>0;i++{Sscanf(Map(func(r rune)rune{if r<'1'{return'1'}else{return'0'}},TrimLeft(Sprintf("%b",n),"0")),"%b",&n)}
return}

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Implements the algorithm more or less verbatim from the OP.

Vyxal 3 s, 3 bytes

½⊻b

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port of the jelly answer

Husk, 6 bytes

ΣẊ≠:0ḋ

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Algorithm from Popov's APL answer.

Husk, 9 bytes

←LU¡ȯḋm¬ḋ

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The approach shown in the question.

Explanation

←LU¡ȯḋm¬ḋ
   ¡      apply function pinfinitely, collecting results
    ȯ     composition of three function
        ḋ convert to binary digits
      m¬  negate each bit
     ḋ    convert back to base-10
  U       split infinite list at first non-unique value
 L        get the length of the list
←         decrement   

C (gcc), 76 Bytes, 72 Bytes (thanks to celingcat)

unsigned n,m,i;f(x){for(i=0;n=x;x^=m-1,i++)for(m=2;n/=2;m+=m);return i;}

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PHP, 50 bytes

<?=array_sum(str_split(decbin($argn&-2^$argn*2)));

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Unfortunately, really don't think I can get it any more golfed than this!

The idea is to count transitions 1->0 or 0->1.

APL, 13 characters/bytes

+/2≠/0,2⊥⍣¯1⊢

base 10 to base 2 conversion with all necessary digits

2⊥⍣¯1⊢

2⊥⍣¯1⊢4812390
---> 1 0 0 1 0 0 1 0 1 1 0 1 1 1 0 0 1 1 0 0 1 1 0

add a leading 0

0,

0,1 0 0 1 0 0 1 0 1 1 0 1 1 1 0 0 1 1 0 0 1 1 0
---> 0 1 0 0 1 0 0 1 0 1 1 0 1 1 1 0 0 1 1 0 0 1 1 0

pair-wise reduction that gives 1 if consecutive elements are not equal

2≠/

2≠/0 1 0 0 1 0 0 1 0 1 1 0 1 1 1 0 0 1 1 0 0 1 1 0
---> 1 1 0 1 1 0 1 1 1 0 1 1 0 0 1 0 1 0 1 0 1 0 1

sum

+/

+/1 1 0 1 1 0 1 1 1 0 1 1 0 0 1 0 1 0 1 0 1 0 1
---> 14

Burlesque, 8 bytes

rib2gnL[

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ri   # Read as integer
b2   # Convert to binary
gn   # Collapse like bits
L[   # Find length

29 bytes

ri0j{j+.j2dg)n!2ug}{^^nz}w!vv

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Actually doing the calculation

ri       #Read val as int
0j       #Push a 0 (counter) to the bottom of the stack
{
 j+.j    # Increment counter and put back
 2dg     # Get the digits of binary val
 )n!     # Not each digit
 2ug     # Convert back to decimal
}
{^^nz}w! # While not zero
vv       # Drop final zero leaving counter

PHP, 55 52 bytes

for($i=$argn;$i;$c++,$i^=(1<<log($i,2)+1)-1);echo$c;

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All done arithmetically/bitwise (no strings). Am stuck getting it smaller than the 51 byte answer though...

Vim, 62 59 bytes

-3 bytes thanks to DJMcMayhem

C0
<C-r>=pri<Tab>'%b',<C-r>")
<Esc>0qqC<C-r>=tr(@",'01','10')
<Esc>:s/^0\+
k<C-a>j@qq@q

Here is the xxd output with the unprintable characters intact:

0000000: 4330 0d12 3d70 7269 0927 2562 272c 1222  C0..=pri.'%b',."
0000010: 290d 1b30 7171 4312 3d74 7228 4022 2c27  )..0qqC.=tr(@",'
0000020: 3031 272c 2731 3027 290d 1b3a 732f 5e30  01','10')..:s/^0
0000030: 5c2b 0d6b 016a 4071 7140 71              \+.k.j@qq@q

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Explanation

C                                   " Delete the number (it goes in @")
0<CR>                               " Print 0 (our counter) and a carriage return
<C-r>=pri<Tab>'%b',<C-r>")<CR><Esc> " Use `printf()` to insert the number as base 2
0qq                                 " Return to column 0, start recording a macro
  C<C-r>=tr(@",'01','10')<CR><Esc>  "   Replace 0s with 1s and vice versa
  :s/^0\+<CR>                       "   Delete leading 0s
  k<C-a>                            "   Increment the number on the line above
  j                                 "   Return to the previous line
  @q                                "   Invoke macro recursively
q@q                                 " Stop recording and invoke macro

Perl 5, 31 + 1 (p) = 32 bytes

$_=(sprintf'%b',$_^$_/2)=~y/1//

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Using @Dennis's method.

Bash, 57 bytes

Packages: Core Utililities, grep, sed, vim (for xxd)

Assume the number is given in binary format. Any length is acceptable :)

xxd -b -c1|cut -d" " -f2|sed s/^0*//|grep -o .|uniq|wc -l

05AB1E, 7 5 bytes

Saved 2 bytes thanks to Dennis.

b0ÛÔg

Without the edge case of 0 this could be 3 bytes bÔg.

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Explanation

b      # convert to binary
 0Û    # trim off leading zeroes
   Ô   # remove adjacent duplicates
    g  # length

Java 7, 64 bytes

long g(Long n){return n.toString(n,2).split("0+").length*2-n%2;}

I know this could be beaten by a port of one of the better answers, but I came up with it in chat, and I can't not post it after Poke said something about it :)

Ruby, 26 bytes

f=->n{n<1?0:-n%4/2+f[n/2]}

Inspired by xnor's Python answer.

Octave, 47 bytes

@(x)(sum(dec2bin(bitxor(x,idivide(x,2)))=='1'))

According to the OEIS entry, the value we are looking for as the solution to this challenge is also equal to the number of 1s in the Gray code for the given integer.

Wikipedia tells me the Gray code can be calculated as x ^ (x >> 1), so in the above function I calculate the Gray code as such, convert it to a binary string, and count how many digits of that string are 1.

CJam, 11 10 bytes

Thanks to @Dennis for saving one byte!

ri_2be`,e&

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Explanation

ri            #e Read as integer
              #e STACK: 97
  _           #e Duplicate
              #e STACK: 97, 97
   2b         #e Convert to binary
              #e STACK: 97, [1 1 0 0 0 0 1]
     e`       #e Run-length encoding
              #e STACK: 97, [[2 1] [4 0] [1 1]]
       ,      #e Length
              #e STACK: 97, 3
        e&    #e Return first value if 0, or else the second value
              #e STACK: 3

Java 7, 71 bytes

int b(Long a){return a==0?0:1+b(~a&-1L>>>64-a.toString(a,2).length());}

I know this is beaten by Geobits' split solution (which will eventually be posted) but this was still fun to write

MATL, 7 bytes

BY'nwa*

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Explanation

          % Implicit input, for example 97
          % STACK: 97
B         % Convert to binary
          % STACK: [1 1 0 0 0 0 1]
 Y'       % Run-length encoding
          % STACK: [1 0 1], [2 4 1]
   n      % Number of elements
          % STACK: [1 0 1], 3
    w     % Swap
          % STACK: 3, [1 0 1]
     a    % Any: gives 1 if any element is nonzero
          % STACK: 3, 1
      *   % Multiply
          % STACK: 3
          % Implicit display

JavaScript (ES6), 26 bytes

f=n=>n&&(n^(n>>=1))%2+f(n)

Works by counting the transitions between 0 and 1. Only works up to 31 bits. 29 bytes to support 53 bits:

f=n=>1<=n&&(n%2^n/2%2)+f(n/2)

PHP, 51 bytes

<?=preg_match_all('/(1+|0+)/',decbin($argv[1])?:o);

Uses a regex to count the number of runs of 1 or 0. Unfortunately this needs a special case for input of 0 that requires 3 additional bytes (and gives a notice).

J, 14 bytes

**1+/@,2~:/\#:

Counts the number of runs in the binary digits of n with the special case returning 0 for n = 0.

Usage

   f =: **1+/@,2~:/\#:
   (,.f"0) 0 1 42 97 170 255 682 8675309 4812390 178956970 2863311530
         0  0
         1  1
        42  6
        97  3
       170  8
       255  1
       682 10
   8675309 11
   4812390 14
 178956970 28
2863311530 32

Explanation

**1+/@,2~:/\#:  Input: integer n
            #:  Get the binary digits of n
       2   \    For each overlapping sublist of size 2
        ~:/       Reduce by not-equals
  1   ,         Prepend a 1
   +/@          Reduce by addition
*               Sign(n), returns 0 for n = 0 else 1
 *              Multiply with the previous sum and return

JavaScript (ES6), 44

Recursive function

Limited to javascript positive integer, 31 bits:

f=(a,s=0)=>a?f((-1>>>Math.clz32(a))-a,s+1):s

Managing double precision number up to 53 significant bits - 59 bytes:

F=(a,s=0)=>a?F('0b'+a.toString(2).replace(/./g,1)-a,s+1):s

In another way: using the amazing algorithm by @Dennis, non recursive function managing up 53 bits, 43 bytes:

a=>a&&a.toString(2).match(/(.)\1*/g).length

Java 7,112 108 100 90 73 bytes

int c(int i){int l=1,j=i;for(;(j=j/2)>0;l*=2);return i<1?0:1+c(2*l-1-i);}

Basic idea

 Lets take an no 10110(21)
 then you do set all bits in no 21 and you will get 11111
 and after that you would subtract the original number from 11111.
 You will get 01001 and loop it until you will get 0

PHP, 64 bytes

based on my countdown solution

for($n=$argv[1];$n;print 1)$n=bindec(strtr(decbin($n),"01",10));

prints 1 character k times, where k is the number of iterations.

+4 bytes for integer output: (empty output for 0)

for($n=$argv[1];$n;$i++)$n=bindec(strtr(decbin($n),"01",10));echo$i;

Racket 349 bytes

(define(h s)(apply string(map(λ(x)(if(eq? x #\0)#\1 #\0))(string->list s))))(define(g s)(let*
((l(string-length s))(k(for/list((i s)(n l)#:final(not(equal? i #\0)))n)))(substring s(last k))))
(define(f n)(if(= 0 n)0(begin(let loop((n n)(c 1))(define m(string->number(string-append "#b"
(g(h(number->string n 2))))))(if(> m 0)(loop m(add1 c))c))))

Ungolfed:

(define (invertBinary s)
  (apply string
         (map
          (λ(x)(if(eq? x #\0)#\1 #\0))
          (string->list s))))

(define (trimLeading0s s)
  (let* ((l (string-length s))
         (k (for/list ((i s)
                       (n l)
                       #:final (not(equal? i #\0)))
              n)))
    (substring s (last k))))

(define (f n)
  (if (= 0 n) 0
      (begin
        (let loop ((n n)
                   (c 1))
          (define m 
            (string->number
             (string-append
              "#b"
              (trimLeading0s
               (invertBinary
                (number->string n 2))))))

          (if (> m 0)
              (loop m (add1 c))
              c)))))

Testing:

(f 0)
(f 1)
(f 42)
(f 97)
(f 170)
(f 255)
(f 682)
(f 8675309)
(f 4812390)
(f 178956970)
(f 2863311530)

Output:

0
1
6
3
8
1
10
11
14
28
32

Haskell, 34 bytes

b 0=0
b n|x<-b$div n 2=x+mod(x+n)2

Python 2, 29 bytes

f=lambda n:n and-n%4/2+f(n/2)

Counts the number of alternations between 0 and 1 in the binary expansion, counting the leading 1 as an alternation. Does so by checking whether the last two binary digits are different, then recursing onto the number with the last digit removed. The last two digits are different exactly if n%4 is 1 or 2, which can be checked as -n%4/2.

Python 2, 30 bytes

lambda n:bin(n^n/2).count('1')

Test it on Ideone.

Background

Let n be a non-negative integer.

Steps 2 and 3 of the process described in the spec can alternatively be stated as removing all leading 1's and toggling the remaining bits.

This means that we'll remove exactly one group of adjacent and equal binary digits in each iteration, so the Binary Countdown Length of n is just the number of these groups in the binary representation of n. For the purposes of this challenge, think of 0 as having no digits.

For n = 8675309, the process looks as follows in binary.

100001000101111111101101
 11110111010000000010010
     1000101111111101101
      111010000000010010
         101111111101101
          10000000010010
           1111111101101
                   10010
                    1101
                      10
                       1
                       0

Instead of counting these groups (which would fail for edge case 0), we do the following.

n and n:2 have the following binary representations.

n   = 8675309 = 100001000101111111101101_2
n:2 = 4337654 =  10000100010111111110110_2

Note that n:2's binary representation is simply n's, shifted one bit to the left.

If we XOR n and n:2, we'll obtain a 1 (MSB), and an additional 1 for each pair of different adjacent digits. The number of groups is thus equal to the number of set bits in n ⊻ n:2.

Jelly, 6 4 bytes

^HBS

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Background

Let n be a non-negative integer.

Steps 2 and 3 of the process described in the spec can alternatively be stated as removing all leading 1's and toggling the remaining bits.

This means that we'll remove exactly one group of adjacent and equal binary digits in each iteration, so the Binary Countdown Length of n is just the number of these groups in the binary representation of n. For the purposes of this challenge, think of 0 as having no digits.

For n = 8675309, the process looks as follows in binary.

100001000101111111101101
 11110111010000000010010
     1000101111111101101
      111010000000010010
         101111111101101
          10000000010010
           1111111101101
                   10010
                    1101
                      10
                       1
                       0

Instead of counting these groups (which would fail for edge case 0), we do the following.

n and n:2 have the following binary representations.

n   = 8675309 = 100001000101111111101101_2
n:2 = 4337654 =  10000100010111111110110_2

Note that n:2's binary representation is simply n's, shifted one bit to the left.

If we XOR n and n:2, we'll obtain a 1 (MSB), and an additional 1 for each pair of different adjacent digits. The number of groups is thus equal to the number of set bits in n ⊻ n:2.

How it works

^HBS  Main link. Argument: n

 H    Halve; yield n:2.
^     XOR n with n:2.
  B   Convert the result to binary.
   S  Compute the sum of the resulting binary digits.

CJam, 14 bytes

ri0{2b:!2bj)}j

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ri      e# read integer
0       e# value for terminal case
{       e# recursive function
  2b    e#   create binary representation with no leading zeros
  :!    e#   flip bits
  2b    e#   convert binary back to integer
  j     e#   recursive call
  )     e#   increment from 0 on the way up
}j      e# end

Basically a knock off of my answer to the other question.