$\newcommand\m{\mathfrak{m}}$
For regular local rings $A$ of dimension $n$ one has the formula:
$$\mathrm{length}(A/\m^{i+1}) = \binom{i+n}{n}.$$
More generally,
let $(A,\m)$ be any Noetherian local ring.
Associated to $A$ is the graded $k = A/\m$-algebra
$$\mathrm{gr}(A) =
\bigoplus_{j=0}^{\infty} \mathrm{gr}^j(A),$$
where $\mathrm{gr}^j(A):=\m^j/\m^{j+1}$. It is a fact that $\mathrm{gr}(A)$
is a local $k$-algebra of dimension equal to $n = \dim(A)$.
The length of $A/\m^{i+1}$ is equal to the sums of the lengths of
$\m^j/\m^{j+1}$ for $j \le i$. Since these latter modules are vector spaces,
this is the same as saying that:
$$\chi_A(i) := \mathrm{length}(A/\m^{i+1}) =
\sum_{j=0}^{i} \dim_k \mathrm{gr}^j(A).$$
The main fact to know is that $\chi_A(i)$
is a polynomial for sufficiently large $i$. This polynomial is known
as the Hilbert-Samuel polynomial; it has degree $n$ and leading coefficient $1/n!$.
If $A$ is regular, however, then $\dim \m/\m^2 = n$, and so by Nakayama's
lemma, $\mathrm{gr}(A)$ is a quotient of $k[x_1,x_2, \ldots, x_n]$.
Since any non-trivial quotient of this has dimension less than $n$, it follows that
$$\mathrm{gr}(A) \simeq k[x_1,x_2, \ldots, x_n].$$
From this it follows that
the Hilbert-Samuel polnomial is exactly:
$$\binom{i+n}{n},$$
and moreover that $\chi_A(i)$ is given by this formula for all $i \ge 0$.
Commutative Artinian rings in general are finite direct products of local Artinian rings, and that has nothing to do with it being an algebra over a special ring.
Every idempotent of such a ring generates an ideal (which is actually a subring) $eRe$. Sometimes it's possible that $e$ splits into two smaller nonzero orthogonal idempotents: $e=f+g$ such that $fg=0$, whereupon $eRe=fRf\oplus gRg$ and the idea splits into two smaller ideals. Using the Artinian condition, you refine the idempotents until they cannot be broken down any more.
The result is a set of finitely many idempotents $e_i$ which cannot be written as a sum of two other orthogonal nontrivial idempotents, and $\sum e_i=1$. The resulting subrings (which are ideals) $e_iRe_i$ are local rings, and $\oplus e_iRe_i=R$.
Best Answer
Your question about simple modules over $A$ can be answered affirmatively. One can see this as follows: take any non-zero $m \in M$ and consider the map of $A$-modules $f:A \to M$ given by $a \mapsto a \cdot m$. This map is non-zero as $1 \cdot m = m \neq 0$ and therefore it must be surjective as $M$ is simple. Therefore we have $M \cong A/\text{ker}(f)$. Next, note that $A/\text{ker}(f)$ can only be simple if $\text{ker}(f) = \mathfrak{m}$: indeed, if $\text{ker}(f) \subsetneq \mathfrak{m}$, then $\mathfrak{m}/\text{ker}(f)$ is a proper submodule of $A/\text{ker}(f)$. We conclude that $M \cong A/\mathfrak{m}$.
For your general question, I don't know if one can say anything more than that $M$ has finite length.