Let $i : R \to S$ be a ring homomorphism (one can view it as inclusion). I'm interested in describing the restriction of scalars of the $S$-module $S^n$. As the rest of the question will demonstrate, I know quite little module theory, so would appreciate answers in "simpler" terms.
Consider a concrete case. Using the notation $\mathbb{Z}_n = \mathbb{Z}/n\mathbb{Z}$, let $i : \mathbb{Z}_2 \to \mathbb{Z}_6$ be the "obvious" inclusion into the $\mathbb{Z}_2$ factor in the CRT decomposition. Equivalently, $i(x) = 3x$. What does the restriction of scalars of $\mathbb{Z}_6^2$ along $i$ look like? Is it a free $\mathbb{Z}_2^2$ module?
My intuition is that it is the same as $\mathbb{Z}_6^2$ as a set, just with the restricted form of multiplication. If this is the case, I struggle to see how to specify a free basis that generates it --- Using the notation $e_i$ for the standard basis, it seems like one cannot just specify $e_1, e_2$. Specifying the elements $a_i e_j$ for $a_i\in\{1,3,5\}, j\in\{1,2\}$ seems like it could work, but this generating set appears to have non-trivial relations.
My motivation for asking this is in some work that wants to use the "matrix algebra" analogy that module theory has (over a commutative ring $S$, with finite free $S$-modules). Specifically one has that $\mathsf{Hom}_S(S^n, S^m) \cong \mathsf{Mat}_{n\times m}(S)$, and there seem to be analogues of things like matrix multiplication, addition, transposition, etc.
These analogues seem to break down if I look at $\mathsf{Hom}_R(R^n, S^m)$ (where $i : R\to S$ is inclusion). The hope is that by an appropriate change of ring (which restriction of scalars seems like the natural choice), I could end up looking at the homomorphism $\mathsf{Hom}_R(R^n, M)$, where $M$ is such that the matrix algebra analogy makes sense. Am I right that such an analogy just doesn't work in this case?