AWK, 179 bytes

func g(x){return gsub(x,X)}{if((d=NF)<2)exit 0
for(;i++<d;)s=$i s
s==$0&&x+=3
(t=5*g("420$"))&&x+=t+1
(t=6*g("69$"))&&x+=t+2
g(800813)&&x+=4
x+=2*g(420)
x+=g(69)
printf"%.1f",x/d}

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Scala, 289 bytes

Port of @97.100.97.109's Python answer in Scala.

Thanks to @Joseph's help to correct the scala code.

Golfed version. Try it online!

s=>BigDecimal((s.sliding(3).count(_=="420")+2*s.sliding(2).count(_=="69")+3*(if(s==s.reverse&s.length>1)1 else 0)+(if(s.contains("800813")) 4 else 0)+5*(if(s.endsWith("420"))1 else 0)+6*(if(s.endsWith("69"))1 else 0)).toDouble/s.length).setScale(1,BigDecimal.RoundingMode.HALF_UP).toDouble

Ungolfed version. Try it online!

object Main extends App {
  
  def calculateScore(s: String): Double = {
    val count420 = s.sliding(3).count(_ == "420")
    val count69 = s.sliding(2).count(_ == "69")
    val palindromeBonus = if (s == s.reverse && s.length > 1) 1 else 0
    val count800813 = if(s.contains("800813")) 4 else 0  // Adjusted to just check if "800813" exists in the string
    val end420Bonus = if (s.endsWith("420")) 1 else 0
    val end69Bonus = if (s.endsWith("69")) 1 else 0

    val totalScore = count420 + 2 * count69 + 3 * palindromeBonus + count800813 + 5 * end420Bonus + 6 * end69Bonus

    BigDecimal(totalScore.toDouble / s.length).setScale(1, BigDecimal.RoundingMode.HALF_UP).toDouble
  }

  override def main(args: Array[String]): Unit = {
    val inputs = List("96942024969", "800813800813", "0", "420", "2435083240587")
    for (i <- inputs) {
      println(s"$i : ${calculateScore(i)}")
    }
  }
}

Python, 136 131 bytes

-5 from @Jonathan Allan

lambda s:"%0.1f"%(((c:=s.count)(m:="420")+2*c("69")+3*(s[:1]<s*c(s[::-1]))+4*("800813"in s)+5*(e:=s.endswith)(m)+6*e("69"))/len(s))

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A crappy solution, but a solution nonetheless.

Vyxal, 302 bits^v2, 37.75 bytes

⁺md69"₌vøEvO?₍λḂ⁼n₀>*;‡800813c$WfÞż∑$L/1∆W

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Many more bytes added to handle single digits not being a palindrome.

Ruby, 114 112 bytes

->n{b=0;[/420/,/69/,n[1]?n.reverse: ?a,/(800813.*)+/,/420$/,/69$/].sum{|x|n.scan(x).size*b+=1.0/n.size}.round 1}

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Excel, 188 bytes

=LET(
    a,800813,
    b,{69,420},
    c,A1,
    d,LEN(c),
    ROUND(
        SUM(
            MMULT(d-LEN(SUBSTITUTE(c,b,"")),1/{1;3}),
            3*(d>1)*(c=0+CONCAT(MID(c,d-SEQUENCE(d)+1,1))),
            4*(FIND(a,c&a)<d),
            (0+RIGHT(c,{2,3})=b)*{6,5}
        )/d,
        1
    )
)

Input in cell A1.

05AB1E, 35 bytes

Ž¥ú2ä©¢IÂQIg≠*I•CΘ=•åI®Å¿)˜ā*OIg/1.ò

Try it online or verify all test cases.

Explanation:

Ž¥ú2ä©¢    # Verify the first two rules:
Ž¥ú        #  Push compressed integer 42069
   2ä      #  Split it into two parts: [420,69]
     ©     #  Store it in variable `®` (without popping)
      ¢    #  Count how many times each occur in the (implicit) input-string
IÂQIg≠*    # Verify the third rule:
I          #  Push the input again
 Â         #  Bifurcate it; short for Duplicate & Reverse copy
  Q        #  Pop both and check if they're the same (aka it's a palindrome)
   Ig      #  Push the input-length
     ≠     #  Check that it's NOT 1 (0 if 1; 1 if >=2)
      *    #  Multiply them together
I•CΘ=•å    # Verify the fourth rule:
I          #  Push the input again
 •CΘ=•     #  Push compressed integer 800813
      å    #  Check if the input contains this number as substring
I®Å¿       # Verify the fifth and sixth rules:
I          #  Push the input yet again
 ®         #  Push pair [420,69] from variable `®`
  Å¿       #  Check whether the input ends with either of these two
)          # Wrap all items on the stack into a list
 ˜         # Flatten the list of pairs and loose values to a single list
  ƶ        # Multiply each value in the list by its 1-based index
   O       # Then sum the list together
    Ig/    # Divide it by the input-length
       1.ò # Round it to 1 decimal after the period
           # (after which the result is output implicitly)

See this 05AB1E tip of mine (section How to compress large integers?) to understand why Ž¥ú is 42069 and •CΘ=• is 800813.

APL(Dyalog Unicode), 65 bytes ^SBCS

1⍕≢÷⍨1⊥4⍸⍤⌽(1<≢×⊢≡⌽),(1∊'800813'⍷,),(,⍤⍉⍤↑'024' '96'(⊃,+/)⍤⍷¨⊂⍤⌽)

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Charcoal, 57 56 55 bytes

﹪%.1f∕⁺⁺ΣＥ⪪42069³⁺×⁺⁵κ¬⊟⪪θι×⊕κ№θι∧∧⊖Ｌθ⁼θ⮌θ³∧№θ800813⁴Ｌθ   

Try it online! Link is to verbose version of code. Explanation:

         Ｅ⪪42069³                                         Map over `420`,`69`
                  ×⁺⁵κ¬⊟⪪θι                               5 or 6 if ends with
                 ⁺                                        Plus
                           ×⊕κ№θι                         1 or 2 times count
                                 ∧∧⊖Ｌθ⁼θ⮌θ³               3 if palindrome
                                           ∧№θ800813⁴     4 if `800813`
      ⁺⁺Σ                                                 Take the sum
                                                    Ｌθ     Divide by length
﹪%.1f                                                       Round to `0.1`
                                                            Implicitly print

Previous 57 56 byte solution:

﹪%.1f∕ΣＥ⟦№θ420№θ69∧⊖Ｌθ⁼θ⮌θ‹⁰№θ800813¬⊟⪪θ420¬⊟⪪θ69⟧×ι⊕κＬθ

Try it online! Link is to verbose version of code. Explanation:

         №θ420                                              Count of `420`s
              №θ69                                          Count of `69`s
                  ∧⊖Ｌθ⁼θ⮌θ                                  Is a palindrome
                          ‹⁰№θ800813                        Includes `800813`
                                    ¬⊟⪪θ420                 Ends with `420`
                                           ¬⊟⪪θ69           Ends with `69`
       Ｅ⟦                                        ⟧          Map over values
                                                  ×ι⊕κ      Multiply by index
      Σ                                                     Take the sum
     ∕                                                Ｌθ    Divide by length
﹪%.1f                                                       Round to `0.1`
                                                            Implicitly print

Retina 0.8.2, 205 201 197 bytes

(?=(.*?420)*)(?=(.*?69)*)(?=((.)+.?(?<-4>\4)+$(?(4)^))?)(?=(.*800813)?).+(?<=(420)?)(?<=(69)?)
$#1$*1$#2$*2$#3$*3$#5$*4$#6$*5$#7$*6/$.&$*
\d
$*
\G1
10$*
/(.+)\1
$1$&
r`.*(\3)*(\3{10})*/(1+)
$#2.$#1

Try it online! Link includes test cases. Explanation:

(?=(.*?420)*)(?=(.*?69)*)(?=((.)+.?(?<-4>\4)+$(?(4)^))?)(?=(.*800813)?).+(?<=(420)?)(?<=(69)?)
$#1$*1$#2$*2$#3$*3$#5$*4$#6$*5$#7$*6/$.&$*

Work out how many times 420 and 69 appear, plus also record whether the input is a nontrivial palindrome, whether it contains 800813, and whether it ends with 420 or 69, plus take its length in unary.

\d
$*

Convert the scores to unary individually; the length is unaffected, since it is already in unary.

\G1
10$*
/(.+)\1
$1$&
r`.*(\3)*(\3{10})*/(1+)
$#2.$#1

Divide the total score by the length, rounding to the nearest 0.1, and converting to decimal.

JavaScript (Node.js), 123 bytes

n=>([s=.5,/420/g,/69/g,n>9&&[...n].reverse().join``,800813,/420$/,/69$/].map((x,i)=>n.replace(x,y=>s+=i*10/n.length))|s)/10

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JavaScript (ES11), 136 bytes

s=>~~([/420/g,/69/g,,800813,/420$/,/69$/].reduce((t,r,i)=>t-~i/3*~~s.match(r)?.length,s>9&[...s].reverse().join``==s)*30/s.length+.5)/10

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Raku, 89 bytes

{round (m:g/420/+2*m:g/69/+3*($_>9&&$_
eq.flip)+4*?/800813/+5*?/420$/+6*?/69$/)/.comb,.1}

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Regex-driven, conveniently.