As a bit of background, consider the category of all covariant, additive functors from a small Abelian category $C$ to Abelian groups, which I'll denote $[C,Ab]$. First, $[C,Ab]$ is an Abelian category, and an easy argument shows that representable functors are projective objects in this category. Indeed, if $T = \text{Hom}_C(X,-)$ is a representable functor and \begin{equation*} 0 \to F_1 \to F_2 \to F_3 \to 0 \end{equation*} is a short exact sequence in $[C,Ab]$ (where exactness is checked on objects of $C$), then Yoneda's lemma gives $\text{Nat}(T,F_i) \cong F_i(X)$, and the result is immediate. I think that the reverse implication also holds (projective implies representable), but I don't remember the proof being as apparent. We even have the Eilenberg-Watts theorems that give criteria for when additive functors (from $R$-Mod to $Ab$) are representable.

Anyway, this is nice, but I find it a bit lacking compared to the tools we have in, say $R$-Mod for some ring $R$. In that setting, we have such results as "A module $P$ is projective iff it is a direct summand of a free module, etc." Or in $Grp$, we have $G$ is projective iff it is free. My point is that we have a notion of "free object" since these are all nice concrete categories, and such a notion seems to have no nice analogue in $[C,Ab]$.

Of course, we have results like $[C,Ab]$ being concrete over $[C,Set]$ which gives a "locally free Abelian group" type example (where $C^{op}$ here would be the category of open sets of a topological space, and $[C,Ab]$ would be presheaves of Abelian groups). However, these are not free objects in $[C,Ab]$.

My question is then "Can $[C,Ab]$ be reasonably thought of as concrete over another category so that we can construct free object?" As an example, is $[R-Mod,Ab]$ concrete over the functor category $[Set,Set]$? (of course, we'd have to juggle Grothendieck universes for this to make any sense; The standard way being to fix some universe $\mathfrak{U}$, and say $Set$ is the category of $\mathfrak{U}$-sets, and let $\mathfrak{U}'$ be the smallest Grothendieck universe containing $\mathfrak{U}$ as an element, so that $Set$ is now $\mathfrak{U}'$-small (see Schuberts "Categories"))