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 $\oplus e_iRe_i=R$.
This question is intimately tied to the idea of dimension. Given a commutative ring $R$, its (Krull) dimension is defined as the supremum of the lengths of chains of all prime ideals.
The name dimension for this quantity makes sense because the coordinate ring of an $n$-dimensional affine variety has Krull dimension $n$. Localization has a geometric interpretation, too: a prime ideal $\mathfrak{p}$ corresponds to an irreducible variety $V$. The localization $R_\mathfrak{p}$ consists of all rational functions that are defined at all points of $V$.
If $R$ is a local ring and its unique maximal ideal $\mathfrak{m}$ is the only prime ideal, then $R$ has dimension $0$, which, if $R$ is Noetherian, is equivalent to being Artinian.
The geometric interpretation above indicates how to produce a local ring with any number of prime ideals. Given a positive integer $n$, consider $n$-dimensional affine space $\mathbb{A}^n$ over a field $k$ of characteristic zero. Its coordinate ring is $k[x_1, \ldots, x_n]$, and if we localize at the prime (maximal, actually) ideal $(x_1, \ldots, x_n)$, we obtain the local ring $k[x_1, \ldots, x_n]_{(x_1, \ldots, x_n)}$ which consists of all rational functions on $\mathbb{A}^n$ that are defined at the origin. By the correspondence theorem for prime ideals in a localization, this ring has a chain of prime ideals
$$
(0) \subsetneq (x_1) \subsetneq (x_1, x_2) \subsetneq \ldots \subsetneq (x_1, x_2, \ldots, x_n)
$$
of length $n$.
As for an example where $R$ is not a domain: an affine variety is irreducible iff its coordinate ring is a domain, so we need to consider reducible varieties. Let $A = \frac{k[x,y]}{(xy)}$ which is the coordinate ring of the union of the lines $x = 0$ and $y = 0$ in the plane. Let $\mathfrak{m} = (x,y)$ and let
$$
R = A_\mathfrak{m} = \left(\frac{k[x,y]}{(xy)}\right)_{(x,y)}
$$
which is the local ring at the origin. Then $(x), (y)$, and $(x,y)$ are all prime ideals of $R$.
Best Answer
First consider the polynomial ring $R=F[x,y]$ for a field $F$. One chain of prime ideals in this ring is $(0)\subseteq (x)\subseteq (x,y)$. Now $(x,y)$ is a maximal ideal, but there are other maximal ideals, for example $(x+1,y)$.
The easiest way to eliminate the other maximal ideals is to pass to the localization at the prime $M=(x,y)$ so that the new ring $R_M$ is a local ring. It is a property of localization that the prime ideals contained in $M$ will have prime counterparts in $R_M$, with the same containment relations and everything. Thus the chain $(0)_M\subseteq (x)_M\subseteq (x,y)_M$ will be a properly ascending chain of prime ideals in the new ring $R_M$.
Even if you are not handy with localization now, you probably will need to be soon, so trying to understand this type of example is worthwhile.