Every Artinian ring is isomorphic to a direct product of Artinian local rings

Yes, your use of the CRT is OK.

It is not hard to show that $R/M^k$ is a local ring for any natural number $k$ and maximal ideal $M$.

The pieces of the ring are Artinian because they are all homomorphic images of the original ring (you just factor out the ideal that is the complement of the factor you're interested in.)

One application of this theorem is that if $n=\prod_{i\in I} p_i^{e_i}$ where $I$ is a finite index set, $p_i$ are distinct primes, and $e_i$ are positive integral exponents, then $\mathbb Z/n\mathbb Z\cong \prod_{i\in I}\mathbb Z/p_i^{e_i}\mathbb Z$.

Another practical corollary is that there are finitely many maximal ideals in such a ring (one for each local summand.)

I'm not sure what other applications you are looking for... It is already quite nice that the ring decomposes into "nicer" ones.