Let $T$ be a site, and let $\mathcal{P}$ denote the category of abelian presheaves on $T$.
Apparently $(Z_U)_{U\in T}$ is a family of generators for $\mathcal{P}$ defined by: $$Z_U(V)=\bigoplus_{\hom(V,U)}\Bbb Z,\quad V\in T,$$
Where it is claimed that: $$F(U)=\hom(\Bbb{Z},F(U))\cong \hom(Z_U,F).$$
I can't see why that last isomorphism holds.
It seems that $\hom(Z_U,F)$ are presheaf homomorphisms, where if I consider any $V$ I have: $$\hom(Z_U(V),F(V))=\hom(\bigoplus_{\hom(V,U)}\Bbb{Z},F(V))\cong \prod_{\hom(V,U)}\hom(\Bbb Z,F(V))\cong F(V)^{\hom(V,U)},$$ and I can't see how considering the commuting squares of the natural transformation gives me information allowing me to conclude.
$Z_U$ is the abelian presheaf freely generated by the set-valued presheaf represented by $U$. Concretely, this says that a map of abelian presheaves out of $Z_U$ is uniquely determined by its action on the standard generators of each direct sum. With this in hand, the desired isomorphism follows from the Yoneda lemma.
Another approach, which I think is a bit nicer, says that an abelian presheaf on $T$ is exactly a presheaf on the preadditive category $\mathbb Z[T]$ freely generated by $T$. This is the preadditive category on the same objects as $T$ with homs the free abelian groups generated by the homs or $T$. The presheaves $Z_U$ become representable in their own right when viewed with domain $\mathbb Z[T]$, and so the claim is the Yoneda lemma outright, just now the version for preadditive categories.