When is a module over $R$ and $S$ an $R \otimes S$-module? Suppose $M$ is a module over $R$ and $S$, commutative rings with $1$.
Under what conditions is $M$ also a $R \otimes S$-module?
Also, a more general question: How to construct a structure of a $R \otimes S$-module? In other words, when one wants to construct a map from $R \otimes S$ to another ring, one can use the universal property of tensor products. What can one do to have a structure of a $R \otimes S$-module?
 A: 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. 
A: For general rings there is a one-to-one correspondence between $R-S$ bimodule structures on $M$ and the $R\otimes S^{op}$ module structures on $M$. 
In your case, when $S$ is commutative, this amounts to an $R\otimes S$ module.  I can't immediately think of an example of a module over two commutative rings which cannot be a bimodule over them... maybe this is already addressed by the QY's post earlier. 
Update:
So, you will be able to get an $R\otimes S$ structure exactly when you can get an $R-S$ bimodule structure. This is indeed more restrictive than merely being a module over the two rings. Mariano Suárez-Alvarez came up with a neat example at An $R$ module and $S$ module that cannot be an $R$-$S$ bimodule demonstrating that it's not always possible to make an $R\otimes S$ module out of $M$. 
A: I think I found a nice way of constructing an $R \otimes S$-module structure.
Notice that the data of an $R$-module structure (for any ring $R$) is given by:


*

*an abelian group $M$, and

*a homomorphism of rings $R \rightarrow End(M)$, where $End(M)$ is the ring of group endomorphisms of $M$ (where the operations are addition and composition).


Having said that, to construct an $R \otimes S$-module structure on $M$, we just need a ring homomorphism from $R \otimes S$ to $End(M)$. This can be constructed using the usual universal property of tensor products.
As an example, let $M$ is an $R$-module and an $S$-module such that the actions of $R$ and $S$ commute. Assume these structures are given by maps $\varphi: R \rightarrow End(M)$ and $\psi: R \rightarrow End(M)$, then we can define a structure of $R \otimes S$-module on $M$ via the ring homomorphism
$r \otimes s \mapsto \varphi(r) \circ \psi(s)$.
