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.)

Let $\{V_i\}$ be representatives from each of the isomorphism classes of simple left $S$-modules. For any finite length 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 the following formula holds (where almost all $\ell(M;V_i)$ are zero because $M$ has finite length, but any single $\ell_R(V_i)$ could be infinite, and $\infty \cdot 0 = 0$):$$\ell_R(M) = \sum \ell_R(V_i) \cdot \ell_S(M; V_i).$$

Thus, a finite length $S$-module $M$ has finite $R$-length if and only if every simple $S$-module that occurs in $M$ has finite $R$-length. This makes it clear that every finite length left $S$-module has finite $R$-length if and only if all simple left $S$-modules have finite $R$-length. The above discussion is true with no requirements on the ring $S$.

Now let's assume that $S$ is finitely generated as an $R$-module. (I'm not sure if this is what you meant by "finitely generated as an $R$-algebra," but it's probably true in the cases that you're studying if you're looking at maximal orders.) I'll prove that $S$ has finitely many simple modules, each of which has finite $R$-length. Let $J(A)$ denote the Jacobson radical of any ring $A$. Then $\mathfrak{m} = J(R) \subseteq J(S)$ because $S$ is a module-finite $R$-algebra; for a proof of this fact, see Lam's A First Course in Noncommutative Rings, Proposition 5.7. Now the simple $S$-modules are the same as the simple $S/J(S)$ modules (see Proposition 4.8 of the same text). But $S/\mathfrak{m} S \twoheadrightarrow S/J(S)$. Because $S$ is module-finite over $R$, $S/\mathfrak{m} S$ has finite $R$-length. Thus $S/J(S)$ has finite $R$-length. So $S/J(S)$ is artinian (the terminology here is that $S$ is a semilocal ring) and thus has finitely many simple modules, each of which has finite $R$-length. The same must be true for the simple $S$-modules

The formula above also verifies your formula in the case of a matrix ring. In case $S = \mathbb{M}_n(R)$, $S$ has a unique simple left module $V = (R/\mathfrak{m} \ \cdots \ R/\mathfrak{m})^T$, the set of all column vectors of length $n$ with entries in $R/\mathfrak{m}$. (At least one way to see that this is the only simple $S$-module is through Morita theory: the Morita equivalence between $R$-$\operatorname{Mod}$ and $S$-$\operatorname{Mod}$ sends the unique simple $R$-module $R/\mathfrak{m}$ to $V$, so that $V$ must be the unique simple $S$-module.) Since $\ell_R(V) = n$, 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.)