I apologize for the text-wall, but I felt like I had to give a pretty detailed framework to properly illustrate the issuses I encountered (scroll to the end to see the final remark/questions):
Let $X$ be a topological space and denote by $\Pi_1(X)$ its fundamental groupoid. Then a local coefficient system $\Gamma$ on $X$ is a functor $$\Gamma \colon \Pi_1(X) \to \textsf{Ab}.$$
From here on we will always assume that our local coefficient systems $\Gamma$ are "constant" on $X$, i.e. that for all $x \in X$ $$\Gamma(x)=A$$ for $A$ a fixed free abelian group.
This is trivially satisfied when $X$ is path connected (by definition of a groupoid).
Define $\mathcal{L}_A$ to be the category with objects $$(X,\Gamma) \in \text{obj}(\mathcal{L}_A),$$ where $X$ is a topological space and $\Gamma$ is a local coefficient system on $X$ (as defined above) and a morphism $$f \in \hom_{\mathcal{L}_A}((X,\Gamma),(Y,G))$$ is a continuous map $f \colon X \to Y$ together with a family of group homomorphisms $$ \forall x \in X \; f_x \colon \Gamma(x) \to G(f(x))$$ such that for all paths $\gamma \colon I \to X$ we have $$G(f \circ \gamma) \circ f_{\gamma(1)}=f_{\gamma(0)} \circ \Gamma(\gamma).$$
We define singular chain complex with local coefficients $$(C_{\bullet}(X;\Gamma),\partial)$$ in the usual way (see for instance Spanier's book or Whiteheads "Elements of Homotopy Theory").
Now let us view this chain complex as a functor on our restricted category, i.e. $$C_{\bullet} \colon \mathcal{L}_A \to \textsf{Comp},$$ where on objects $(X,\Gamma)$ the $C_{\bullet}(X,\Gamma)$ simply is, as the notation suggests, the singular chain complex on $X$ with local coefficients in $\Gamma$. For a morphism $f$ (as above) and a $k$-chain $a \, \sigma$ the functor operates as follows: $$C_k(f)a \, \sigma=f_{\sigma(e_0)}(a) \, (f \circ \sigma).$$
I tried to find a family of models $$\mathcal{M}_i \subseteq \text{obj}(\mathcal{L}_A)$$ such that the functor $C_i$ is free with basis in $\mathcal{M}_i$:
Freeness is always for free (no pun intended) by definition of the chain complex and our choice of $A$ (recall, it is a free abelian group!). By "mimicking" the standard construction for models in the usual singular chain setting, one is naturally led to do the following:
Fix a basis $\{a\}$ of $A$ and define $$\mathcal{M}_i=\{M_{(i,a)} \mid a \in \{a\}\}, \; \text{where} \; M_{(i,a)}=(\Delta^i,A).$$ (Here by $A$ we abusively mean the trivial local coefficient system on $\Delta^i$). For the $C_i$-model set we could try $$\{a \, l_i \mid a \in \{a\}\},$$ where $$l_i \colon \Delta^i \to \Delta^i$$ is the identity.
To check whether our choices are correct we need to fix an object $(X,\Gamma)$ in $\mathcal{L}_A$ and show that the set $$\{C_i(f)a \, l_i \mid f \in \hom_{\mathcal{L}_A}(M_{(i,a)},(X,\Gamma)), \, a \in \{a\}\}$$ defines a basis of $C_i(X,\Gamma)$. It is not hard to show that the standard basis of $C_i(X,\Gamma)$ is contained in the set above, however the other inclusion does not seem to hold:
We can choose some continuous $f \colon \Delta^i \to X$ and set $f_s=0$ for all $s \in \Delta^i$. This defines a morphism in $\mathcal{L}_A$, but then $$C_i(f)a \, l_i= 0 \; f=0 \in C_i(X,\Gamma)$$ and the $0$ element can't possibly lie in any nontrivial basis.
Questions: Is my argument flawed, i.e. am I actually wrong somewhere and this choice of models does actually work? Alternatively, does someone see a possibility or another choice in order to get a familiy of models?
I thought about restricting the set of homomorphisms $f$ in a way that $f_s$ maps the fixed basis $\{a\}$ into itself, which seems very restrictive.
Thanks!