I will assume our algebra to have an identity.
Question 1. How about: a representation is decomposable if and only if there exist two non-zero idempotent matrices $A_1$ and $A_2$ such that
- both commute with all elements of the representation,
- $A_1A_2=A_2A_1=0$ and
- $A_1+A_2=I$, the identity.
If a representation is decomposable, then clearly such matrices exist. Conversely, the first condition implies that the images of $A_1$ and of $A_2$ are subrepresentations, the second one implies that they intersect trivially$^1$, and the third one implies that their sum (and therefore their direct sum) is the whole representation.
Question 2. Let $A$ be an algebra. The following are equivalent:
- a representation of $A$ is irreducible if and only if it is indecomposable;
- every representation is a direct sum of irreducibles;
- $A$ is Artinian and as a (left, say) module over itself is a direct sum of irreducibles;
- $A$ is isomorphic to a direct sum of matrix algebras over division algebras.
The last equivalence is the Artin-Wedderburn theorem, and it completely classifies the situation you are asking about. An important example of such algebras is given by group algebras $K[G]$, where $G$ is a finite group, and $K$ is a field of characteristic not dividing $|G|$.
$^1$ If $v$ is in the image of $A_2$, then $A_2v=v$, since $A_2$ is idempotent, so $A_1v=0$ by the second condition; so if $v$ is also in the image of $A_1$, then by the same argument $v=A_1v=0$.