I think I've gotten about halfway there but I can't finish it out, so maybe this can help someone else.
As you said, you've given the individual parts of the scores, there are three parts.
#1
Seems to based on the length of the word, $\ell$. If $\ell \equiv 0 \ (\text{mod} \ 2)$, the first score is $0.01\ell$. Otherwise, the first score is $0.02\ell$. This can be written concisely using the modulo operator $\%$ as $$0.01\ell(\ell\%2 + 1)$$
#2
Seems to be based on the numerical value of the initial letter. Define $\#(\alpha)$ as the numerical value of a letter $\alpha$,so $\#(a) = 1, \ \#(b) = 2, \dots, \#(z) = 26$. Notably, the second score is a strictly increasing function of $\#$ of the first letter in each word. Even more notably, for each $\alpha$ such that $\#(\alpha) \equiv 1 \ (\text{mod} \ 3)$ we have the second score is $\frac{1}{300}(1+23\#(\alpha))$. We only have one $\#(\alpha) \equiv 0 \ (\text{mod} \ 3)$, but we do have that for that one, $\frac{1}{300}(23\#(\alpha))$, as a linear function, this goes through the origin if we graph it, so it seems reasonable. However, the pattern doesn't continue. If we make a third parallel line for $\#(\alpha) \equiv 2 \ (\text{mod} \ 3)$, we get $\frac{1}{300}(44+23\#(\alpha))$, which is nice and round, but I was hoping it would be a $2$ instead of a $44$. If that were the case, we would have been able to write the second score as $$\frac{1}{300}(\#(\alpha)\%3+23\#(\alpha))$$ but this isn't the case. I could be way off on this one, but the fact that 5 points are colinear and have the same remainder seems too much a coincidence.
#3
I don't really have a clue about this one. It's generally increasing with respect to the length of the words, but not exactly. It's almost strictly increasing with respect to the sum of the $\#$ values for all the letters in the word, but "hunger" and "seagull" are flipped. It could be some variation of this, where maybe vowels and consonants are worth different values, but I couldn't find anything.
And lastly,
If what I have so far is correct, then "voldemortswrath" should be 0.3 + 1.69 + 7.35 = 9.34, further suggesting that the third score is somehow related to the word length/numerical values.
voldemortswrath = 9.34
! $\endgroup$