C#6, 211 bytes

using static System.Math;(h,f,x,y)=>{int b=h>5?24/(50/h)*100:Min(f%2<1?(int)Ceiling(f*.1)<<h+2:10<<h,200),c=-(int)Ceiling(b*.1*(y?2:x?6:4))*100,d=-(int)Ceiling(b*.1*(y?x?2:1:0))*100;return new[]{-c-d*2,c,d,d};};

Contains an idea shamelessly stolen from @Sp3000.

Assign to Func<int,int,bool,bool,int[]>
h=han
f=fu
x=true if dealer wins, false if non-dealer wins
y=true if self draw, false if discard
returns int[], [0]=winner, [1]-[3]:loser
repl.it demo

Ungolfed + explanations

// b is base score / 10
int b=h>5
    // han is [6,13], use Sp3000's idea
    ?24/(50/h)*100
    // han <= 5, calculate base score (/ 10) from formula...
    // Yes this formula works for han==5 also
    :Min(
        f%2<1
            // round up to 10, then times 2^(h+2)
            ?(int)Ceiling(f*.1)<<h+2
            // b=2.5*2^(h+2)=10*2^h=10<<h
            :10<<h
        // Cap at 200, thus base score is capped at 2000
        ,200),
// Multiply 0.1b by 2/4/6 according to input, then multiply by 100 to get actual score
c=-(int)Ceiling(b*.1*(y?2:x?6:4))*100,
d=-(int)Ceiling(b*.1*(y?x?2:1:0))*100;
// Array type implied from arguments
return new[]{-c-d*2,c,d,d};

Python 2, 128 bytes

def m(h,f,o,t):C=100;b=[24/(50/h)*1000,min(2000,[-f/10*~9,f][f%2]<<h+2)][h<5];x=(o|2-t)*-2*b/C*C;y=~o*t*b/C*C;return-x-y-y,x,y,y

h is han, f is fu, o is dealer (1/0), t is self-draw (1/0). Outputs in the same order as the test cases. ideone link.

Expanded slightly:

def m(h,f,o,t):
  C = 100

  # Base score
  b = [24/(50/h)*1000, min(2000,[-f/10*~9,f][f%2]<<h+2)][h<5]

  # Player scores - uses the -x/100*-100 approach to round up
  x = (o|2-t)*-2*b/C*C
  y = ~o*t*b/C*C

  return -x-y-y, x, y, y

Indirect thanks to @MartinEnder.

05AB1E, 79 bytes

05AB1E uses CP-1252 encoding.

Input order:

Han
Fu
Dealer = 1, Non-dealer = 0
Self = 2, Discard = 0

Code:

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Try it online!

Explanation to come.