To answer your first question, one way to define (co)homology with local coefficients is the following.
Let $X$ be a space, let $\Pi(X)$ be the fundamental groupoid of $X$, ie. a category with objects points of $X$ and morphisms $x \rightarrow y$ given by homotopy classes of paths. A local coefficient system $M$ on $X$ is a functor $M: \Pi(X) \rightarrow \mathcal{A}b$ from the fundamental groupoid to the category of abelian groups. In particular, $M$ associates a "group of coefficients" $M(x)$ to every point of $x \in X$.
Associated to $M$ is the singular complex given by
$C _{n}(X, M)= \bigoplus _{\sigma \in Sing_{n}(X)} M(\sigma(1,0,\ldots,0))$,
where $Sing _{n}(X)$ is the set of all maps $\Delta ^{n} \rightarrow X$. The differential can be defined using the fact that the groups $M(x)$ are functorial with respect to paths in $X$. Observe that this is a very similar to the usual definition of singular complex with coefficients in an abelian group $A$, which would be
$C _{n}(X, A) = \bigoplus _{\sigma \in Sing_{n}(X)} A$,
except in the "non-local" case, we count the occurances of any $\sigma: \Delta^{n} \rightarrow X$ in a given chain using the same group $A$ and in the local case, we use $M(\Delta^{n}(1, 0, \ldots, ))$, which might be different for different $\sigma$.
This is the locality (localness?) in the name, which should be contrasted with globality of usual homology with coefficients, where the choice of the group $A$ is global and the same for all points.
The definition I have given above is enlightening but perhaps not suitable for computations. Luckily under rather weak assumptions one can use the definition you allude to. Let me explain. Let $X$ be path-connected, let $x \in X$ and consider $\pi = \pi_{1}(X, x)$ as a category with one object and morphisms the elements of the group.
The obvious inclusion $\pi \hookrightarrow \Pi(X)$ is an equivalence of categories and so the functor categories $[\pi, \mathcal{A}b], [\Pi(X), \mathcal{A}b]$ are equivalent, too. But the left functor category is exactly the category of $Z[\pi]$-modules! In particular, we have a bijection between isomorphism classes of local coefficient systems on $X$ and $Z[\pi]$-modules.
If $X$ is nice enough to admit a universal cover $\tilde{X}$, the above allows us to give another definiton of homology with local coefficients, the one you know. Let $M$ be a local coefficient system and let $M^\prime$ be the associated $\mathbb{Z}[\pi]$-module under the above equivalence (which is unique up to a unique isomorphism). Since $\pi$ acts on $\tilde{X}$, it also acts on $C _{\bullet}(X, \mathbb{Z})$ and so the latter is a chain complex of $\mathbb{Z}[\pi]$-modules. We the can define homology with local coefficients to be homology of the complex
$C_{n}(X, M ^\prime) = C_{n}(\tilde{X}, \mathbb{Z}) \otimes _{\mathbb{Z}[\pi]} M^\prime$
One can show that this two definitions I gave agree, that is, for $X$ like above we have an isomorphism $H_{n}(X, M) \simeq H_{n}(X, M^\prime)$.
Best Answer
If I understand correctly, here is a clearer rephrasing of your question. Suppose $R$ is a PID, $X$ is a chain complex of free $R$-modules, and $S$ is an $R$-algebra. Using the universal coefficients theorem, you can compute the homology $$H_n(X; S) \cong H_n(X; R) \otimes_R S \oplus \text{Tor}_1^R(H_{n-1}(X; R), S)$$ as an $R$-module. However, $H_n(X;S)$ is not just an $R$-module but an $S$-module, and you are pointing out that the universal coefficients theorem does not tell you what $H_n(X;S)$ is as an $S$-module.
You are correct that the universal coefficients theorem as usually stated does not tell you the $S$-module structure. However, you actually still can figure out the $S$-module structure from it. First of all, let me remark that in many cases, the $S$-module structure is actually automatically uniquely determined by the $R$-module structure. For instance, if $S$ is a quotient or localization of $R$, then any $R$-module has at most one $S$-module structure. This in particular covers the usual cases where $R=\mathbb{Z}$ and $S$ is $\mathbb{Z}/(n)$ or $\mathbb{Q}$.
Even when the $S$-module structure is not automatically determined, though, you can still figure it out by the naturality of the universal coefficients theorem. Specifically, the short exact sequence $$0\to H_n(X; R) \otimes_R S \to H_n(X; S) \to \text{Tor}_1^R(H_{n-1}(X; R), S)\to 0$$ is natural in the coefficient module $S$. For any $s\in S$, there is an $R$-module homomorphism $S\to S$ given by (right) multiplication by $S$, and naturality of the above sequence with respect to these homomorphisms says exactly that the maps in the sequence above are homomorphisms of (right) $S$-modules, not just of $R$-modules. Moreover, if you examine the proof of the universal coefficients theorem, you can actually choose the splitting of this exact sequence to also be natural in $S$, for any fixed chain complex $X$ (the splitting comes from a choice of splitting of the inclusion of the $n$-cycles into $X_n$, and once you fix that splitting everything else is natural). So, the isomorphism $$H_n(X; S) \cong H_n(X; R) \otimes_R S \oplus \text{Tor}_1^R(H_{n-1}(X; R), S)$$ is actually an isomorphism of $S$-modules, not just of $R$-modules.