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.
No, the algebraic Kunneth sequence tells us that if we have free chain complexes C,D over the PID R, then we have the short exact sequence:
$0 \rightarrow \bigoplus\limits_{i+j=k} H_i(C) \otimes_R H_j(D) \rightarrow H_k (C \otimes D) \rightarrow \bigoplus\limits_{i+j=k-1} Tor^R_1 (H_i(C),H_j(D)) \rightarrow 0$.
The Eilenberg-Zilber theorem tells us that $S_*(X \times Y,R)$ and $S_*(X,R) \otimes S_*(Y,R)$ are quasi-isomorphic. From these one can deduce the usual Kunneth formula in topology. To figure out the case for coefficients in a module M we can just tensor each of these free chain complexes by $M$ and we preserve the quasi-isomorphism. So we have $S_*(X \times Y,R) \otimes_R M$ is quasi-isomorphic to $S_*(X,R) \otimes S_*(Y,R) \otimes_R M$. Rewriting, the former is $S_*(X \times Y,M)$ and the latter is $S_*(X,R) \otimes_R S_*(Y,M)$.
Applying the algebraic Kunneth sequence tells us that we have a short exact sequence:
$0 \rightarrow \bigoplus\limits_{i+j=k} H_i(X,R) \otimes_R H_j(Y,M) \rightarrow H_k (X \times Y,M) \rightarrow \bigoplus\limits_{i+j=k-1} Tor^R_1 (H_i(X,R),H_j(Y,M)) \rightarrow 0$
And of course you can swap the roles of X and Y if you'd like to pick which space gets which coefficients. The reason why your proposed sequence does not work is because there are modules such that $M \otimes_R M \neq M$.
In the case R is not a PID, you can use the algebraic Kunneth spectral sequence because I believe the Eilenberg-Zilber theorem holds with no conditions on the ring. This spectral sequence involves the higher Tor terms which vanish for PID's.
Best Answer
Look at $R=\mathbb Z, N = \mathbb Z/2$.
Then you have the two short exact sequences $0\to H_2(\mathbb RP^2; \mathbb Z)\otimes \mathbb Z/2\to H_2(\mathbb RP^2;\mathbb Z/2)\to \mathrm{Tor}_1^\mathbb Z(H_1(\mathbb RP^2;\mathbb Z), \mathbb Z/2)\to 0$
and
$0\to H_2(S^2; \mathbb Z)\otimes \mathbb Z/2\to H_2(S^2;\mathbb Z/2)\to \mathrm{Tor}_1^\mathbb Z(H_1(S^2;\mathbb Z), \mathbb Z/2)\to 0$
In the first exact sequence, the leftmost term is $0$, and in the second one it's the rightmost term which is $0$.
If the splitting were natural, since $H_2(\mathbb RP^2; \mathbb Z)\otimes \mathbb Z/2\to H_2(S^2; \mathbb Z)\otimes \mathbb Z/2$ and $\mathrm{Tor}_1^\mathbb Z(H_1(\mathbb RP^2;\mathbb Z), \mathbb Z/2)\to \mathrm{Tor}_1^\mathbb Z(H_1(S^2;\mathbb Z), \mathbb Z/2)$ are both $0$ (because in each one, one of the terms is $0$), then $H_2(\mathbb RP^2;\mathbb Z/2)\to H_2(S^2;\mathbb Z/2)$ would be $0$ (can you see how this relates to naturality of the splitting ?)
But it's not $0$ (I don't know what's the easiest way to see this, here are at least two possibilities :
a) Use Poincaré duality and try to convince yourself that your map sends the fundamental class (mod $2$) $[\mathbb RP^2]$ to $[S^2]$
b) Use the following cofiber sequence : $S^1\overset 2\to S^1\to \mathbb RP^2$ and note that it continues as $S^1\to S^1\to \mathbb RP^2\to S^2\to S^2$ and that the map $\mathbb RP^2\to S^2$ is the one you're looking for. This implies that the map $S^2\to S^2$ is $\Sigma 2 = 2 : S^2\to S^2$ so it induces $0$ in mod $2$ homology, and so the long exact sequence in homology of the cofiber sequence $\mathbb RP^2\to S^2\to S^2$ shows that the first map is an iso on $H_2(-;\mathbb Z/2)$ )