On page 11 in Tensor Categories is the statement
Let $\mathcal{C}$ be a finite abelian $k$-linear category. Then for any $X,Y \in \mathcal{C}$ with $X$ simple we have \begin{align*} \dim_k \mathcal{C}(P(X), Y) = [Y:X] \ , \end{align*} where $P(X)$ is the projective cover of $X$ and $[Y:X]$ is the multiplicity of $X$ in the composition series of $Y$.
Why is this true?
Assuming$^1$ $\dim_k \mathcal{C}(P(X), \tilde{X}) = \delta_{X,\tilde{X}}$ for any two simples $X,\tilde{X}$, I can at least produce a well-defined map from $[Y:X]$ to $\mathcal{C}(P(X),Y)$. I don't know how to take it from there.
$^1$ but I also don't know if this is true. Are projective covers of simples indecomposable?
All of your favorite facts about projective covers from modules over finite-dimensional algebras carry over to finite abelian linear categories; indecomposable projectives are the projective covers of simple objects, unique decomposition, etc. With this in mind, we can prove the formula by induction on the length of $Y$.
First suppose $Y$ is simple. Any nonzero morphism $P(X) \to Y$ is an epimorphism, so if such a morphism exists then $P(Y)$ is a direct summand of $P(X)$. But both $P(Y)$ and $P(X)$ are indecomposable, so in this case $P(Y) = P(X)$ and thus $Y = X$. Hence $\text{Hom}(P(X),Y) = 0$ if $Y \neq X$. Now, a general fact about projective covers is that the natural map $\text{End}(X) \to \text{Hom}(P(X),X)$ induced by $p_X : P(X) \to X$ is a vector space isomorphism. This follows from the fact that $\ker p_X$ is superfluous. If $k$ is algebraically closed (this is an assumption that the authors make throughout the text), then $\text{End}(X) = k$ and hence $\dim \text{Hom}(P(X),X) = 1$. We thus have $$\dim \text{Hom}(P(X),Y) = \delta(X,Y) = [Y:X].$$
Now for the inductive step. Assume $Y$ has length $\ge 2$ and let $Z$ be a simple subobject of $Y$. Since $P(X)$ is projective, the exact sequence $$0 \to Z \to Y \to Y/Z \to 0$$ induces an exact sequence $$0 \to \text{Hom}(P(X),Z) \to \text{Hom}(P(X),Y) \to \text{Hom}(P(X),Y/Z) \to 0.$$ By induction, $\dim \text{Hom}(P(X),Z) = [Z:X]$ and $\dim \text{Hom}(P(X),Y/Z) = [Y/Z:X]$. Therefore $$\dim \text{Hom}(P(X),Y) = [Z:X] + [Y/Z:X] = [Y:X].$$