To answer your first question, every finite length $S$-module also has finite $R$-length.
Suppose that $S$ is free of rank $n$, so that $S \cong R^n$ as $R$-modules. Then by additivity of the length of $R$-modules, $\ell_R(S) = n \ell_R(R) < \infty$. Notice that, because $R$ is artinian, $S$ is artinian; thus for $S$-modules, finitely generated is the same as finite length. Now for any finitely generated left $S$-module $M$, the fact that there is some surjective homomorphism $S^d \twoheadrightarrow M$ implies that $$\ell_R(M) \leq \ell_R(S^d) = d \cdot \ell_R(S) < \infty$$.
The second question, about a general formula relating the two lengths, is a bit more tricky. But I think we can say something.
Let $V_1, \dots V_r$ be representatives from each of the isomorphism classes of simple left $S$-modules. For any finitely generated module ${}_S M$, let $\ell_S(M; V_i)$ denote the number of times that $V_i$ occurs in a composition series for $M$. Then we can say the following:$$\ell_R(M) = \sum \ell_R(V_i) \cdot \ell_S(M; V_i)$$.
In particular, this verifies your formula in the case of a matrix ring. In case $S = \mathbb{M}_n(R)$, $S$ has a unique simple left module $V = (R/m \ \cdots \ R/m)^T$, the set of all column vectors of length $n$ with entries in $R/m$. Since $r = 1$ and $\ell_R(V) = n$, the formula above reduces to $\ell_R(M) = n \cdot \ell_S(M)$.
(If something above doesn't make sense, you may want to review Jordan-Hoelder theory.)