Commutativity is unimportant. If $R, S$ are two rings and $M$ is a (left) module over both of them, then all you know is that $M$ is a (left) module over the free product $R*S$ by the universal property of the free product (a slight misnomer as the free product is actually the coproduct in the category of rings). The free product is very different from the tensor product, and among other things is often noncommutative even if $R, S$ are commutative; for example the free product of $\mathbb{Z}[x]$ with itself is $\mathbb{Z} \langle x, y \rangle$, the ring of noncommutative polynomials in two variables (e.g. the free ring on two elements).
To get an $R \otimes S$-module structure you need to assume that the actions of $R$ and $S$ commute, i.e. that $r(s(m)) = s(r(m))$ for all $r \in R, s \in S$.
To get $R \otimes S$-modules in general, take a tensor product $M \otimes N$ where $M$ is an $R$-module and $N$ is an $S$-module.