Say that we have a morphism of commutative rings $f: R \to S$. Does the restriction of scalars functor $f^*: S \text{Mod} \to R \text{Mod}$ commute with tensor products? In other words, I would like to know if we have an isomorphism of $R$-modules: $$f^*(V \otimes_S V') \cong f^*V \otimes_R f^*V'.$$ If it does not hold in general, then under what conditions is it true?
The reason I am wondering is that it seems that an $S$-bilinear map $\phi: V \times V' \to V''$ determines an $R$-bilinear map under restriction of scalars.
I get the feeling that it is not quite right to say that $f^*$ is a monoidal functor, but I am not sure what the correct relationship to tensor products should be.
I will denote the restriction by $M \mapsto M|_R$. We always have a canonical homomorphism of $R$-modules $$\theta_{M,N} : M|_R \otimes_R N|_R \to (M \otimes_S N)|_R,\quad m \otimes n \mapsto m \otimes n$$ where $M,N$ are two $S$-modules. This is because the map $$M|_R \times N|_R \to (M \otimes_S N)|_R,\quad (m,n) \mapsto m \otimes n$$ is $R$-bilinear, so that we can use the universal property of the tensor product (that is, its definition). There is also a natural homomorphism of $R$-modules $R \to S|_R$. These maps endow $M \mapsto M|_R$ with the structure of a lax monoidal functor $(\mathbf{Mod}_S,\otimes_S) \to (\mathbf{Mod}_R,\otimes_R)$. This amounts to saying that the maps above are natural (in the sense of category theory, i.e. they are components of a natural transformation) and compatible with each other in a certain way.
In general, it is not a strong monoidal functor. This happens only when $R \to S$ is an isomorphism of rings (since for a strong monoidal functor, also the unit map needs to be an isomorphism). But your question is only about the homomorphisms $\theta_{M,N}$.
Notice that $\theta_{M,N}$ is the component of a natural transformation of two functors $\mathbf{Mod}_S \times \mathbf{Mod}_S \to \mathbf{Mod}_R$, both of which are cocontinuous in each variable. This is because scalar restriction is a cocontinuous functor and the tensor product is cocontinuous in each variable (in fact, both of these functors are left adjoints). Now, since $S$ generates $\mathbf{Mod}_S$ under colimits, it follows formally that $\theta_{M,N}$ is an isomorphism for all $M,N \in \mathbf{Mod}_S$ if and only if this is the case for $M = N = S$. In this case, the map $\theta_{S,S}$ can be identified with the multiplication homomorphism $$S \otimes_R S \to S, \quad a \otimes b \mapsto ab,$$ which of course does not have to be an isomorphism (for example, when $R$ is a field, this is the case iff $\dim_R(S) \leq 1$).
Actually, this map is an isomorphism if and only if $R \to S$ is an epimorphism in the category of commutative rings (since, in general, a morphism $X \to Y$ in a category with pushouts is an epimorphism iff the codiagonal $Y \sqcup_X Y \to Y$ is an isomorphism). Notice that epimorphisms of commutative rings don't have to be surjective (for example, every localization is an epimorphism). More on the structure of epimorphisms can be found in
Appendix. There is a more general observation which allows you to get the lax monoidal structure on $M \mapsto M|_R$ if you already know the strong monoidal structure on $N \mapsto N \otimes_R S$, its left adjoint. Namely, if $F : \mathcal{C} \to \mathcal{D}$ is a strong monoidal functor (actually, oplax suffices) and $G : \mathcal{D} \to \mathcal{D}$ is right adjoint to $F$ with unit $\eta$ and counit $\varepsilon$, then $G$ carries the structure of a lax monoidal functor: We take $$G(X) \otimes G(Y) \xrightarrow{\eta} G(F(G(X) \otimes G(Y))) \to G(F(G(X)) \otimes F(G(Y)) ) \xrightarrow{\varepsilon \otimes \varepsilon} G(X \otimes Y)$$ and $$1_{\mathcal{C}} \xrightarrow{\eta} G(F(1_{\mathcal{C}})) \to G(1_{\mathcal{D}}).$$