(My original answer contained a significant error, which is what the comments below refer to.)
Yes, it is true. However, it is not true if the collection of modules is infinite. User26857 gives an example in the comments.
Lemma. Let $M_i$ , $i \in I$, be a finite collection of modules. Then the natural map $$\phi : (\prod M_i) \otimes N \to \prod ( M_i \otimes N)$$ is an isomorphism. (The natural map is the one induced by the universal property of $\prod (M_i \otimes N)$, and $\_ \otimes N$ applied to the maps $\prod M_i \to M_i$.)
Proof. This is because finite products are finite coproducts, and tensor products commute with coproducts.
If $I$ is infinite, then this map is neither injective nor surjective in general.
We can use user26857s example to produce an example when $\phi$ is not injective: $(\prod_{n \geq 0} K[x]/x^n) \otimes K(x) \to \prod_{n \geq 0} (K[x]/x^n \otimes K(x)) = 0$. The LHS is not $0$ because the map $K[x] \to \prod_{n \geq 0} K[x]/x^n$ is injective, so because $K(x)$ is flat, $K(x)$ embeds as a submodule of $(\prod_{n \geq 0} K[x]/x^n) \otimes K(x)$.
An example when $\phi$ is not surjective is $(\prod_{\mathbb{N}} \mathbb{Z}) \otimes \mathbb{Q} \to \prod_{\mathbb{N}} \mathbb{Q}$. The element $(1,1/2,1/4,1/8, \ldots)$ is not in the image.
Proposition. Lei $I$ be finite again. Suppose $M_i$ are submodules of $M$. We identify each $M_i \otimes N$ with a submodule of $M \otimes N$, using that $N$ is flat. We also identify $(\bigcap_{i \in I} M_i) \otimes N$ with a submodule of $N$ in the same way. Then $\bigcap_{i \in I} (M_i \otimes N) = (\bigcap_{i \in I} M_i) \otimes N$, as submodules of $M$.
Proof.
Consider the exact sequence:
$$0 \to \bigcap_{i \in I} M_i \to M \to \prod_{i \in I} M / M_i.$$
Using flatness of $N$, we get another exact sequence:
$$0 \to (\bigcap_{i \in I} M_i) \otimes N \to M \otimes N \to (\prod_{i \in I} M / M_i) \otimes N\qquad (**)$$
We study the map $f: M \otimes N \to (\prod_{i \in I} M / M_i) \otimes N$ by composing it with $\phi: (\prod_{i \in I} M / M_i) \otimes N \to \prod_{i \in I} ((M / M_i) \otimes N)$ from the lemma above.
Claims:
1) The composition $\phi \circ f$ is the natural projection map. Hence the kernel of $\phi \circ f$ is $\bigcap_{i \in I} (M_i \otimes N)$. (Lazy justification -- all maps are induced canonically...)
2) $\ker(\phi \circ f) = \ker(f)$. This is because $\phi$ is injective, as we saw in the lemma.
Now it follows from exactness of the sequence $(**)$ above that, as submodules, $(\bigcap_{i \in I} M_i) \otimes N$ and $\ker f$ agree. This proves the claim.
No, this is usually not true. For instance, suppose that $S$ is actually a subring of $R$, and the action of $S$ on $M$ is just given by restricting the action of $R$. Then $M$ will not be free over $R\otimes_\mathbb{Z} S$ (unless $S$ is just $\mathbb{Z}$), since for each $s\in S$, $s\otimes 1$ and $1\otimes s$ act on $M$ in the same way.
I don't know of any reasonable conditions that would make the answer yes. Certainly it's not enough to just put conditions on $R$ and $S$ themselves, since $R$ and $S$ could be arbitrarily nice in the example above (for instance, $R$ could a number field and $S$ could be its ring of integers). What you need is for the actions of $R$ and $S$ on $M$ to be "independent" in some strong way.
Best Answer
No, because you are only inverting the polynomials in $x$. e.g.
$$ k(x) \otimes_{k[x]} k[x,y] \cong k(x)[y]$$
Also,
$$ k(x) \otimes_{k[x]} k[x] / (x) \cong 0 $$