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. The conditions "irreducible" and "indecomposable" for representations of an algebra are equivalent if and only if every representation is a direct sum of irreducibles if and only if the algebra is Artinian and as a (left, say) module over itself is a direct sum of irreducibles if and only if the algebra is a direct sum of matrix algebras over division algebras. The last statement 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 $v=A_1v=0$.