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 $v=A_1v=0$.