Well, as you said, reducible representations are made up of irreducible ones. That is
always the case (not under special circumstances, as you say) though in some circumstances it may be more
obvious.
But consider for example this case: you have an irreducible representation with
nx
n matrices
M. Then you can make another representation of
k n x
k n by making new matrices, which have
k times
M on their diagonal. Clearly, this representation doesn't add any new physical information, it's just bigger and therefore more unpleasant to work with.
Similarly, all the information in a representation is stored in its irreducible parts, which are always smaller (or as large, but never bigger) than the reducible representation. Therefore, to understand the physics, it's much easier to look at the smaller blocks and understand each of them separately. Then a reducible representation contains just the same information. I also think that there are only finitely many irreducible representations (up to some obvious isomorphisms, maybe) while there are infinitely many irreducible ones (in the previous paragraph, there is a recipe to make infinitely many from just one

)
Also, view the converse situation. You are doing a calculation, and out comes some representation in big matrices. It is a hell of a job to find all the physics in there. But it may turn out, that you can split it into smaller parts, of which you know exactly what they do, and you actually already know everything that makes up the representation. From this point of view, it's something like diagonalizing a matrix to read off the eigenvalues: if you have the right diagonalization technique its easier than solving the characteristic polynomial - and the matrix may look very complicated, but it will become extremely easy if you choose the right basis (OK, so maybe that was a bad analogy, but it was the first to come to mind and it seems appropriate at first sight

).
By the way, mathematicians do the same. Give a mathematician some large group, and the first thing (s)he'll do is try to find all the (normal) subgroups to see if it's made up of pieces that (s)he already knows.
[edit]Dr Transport said it just right, and much shorter

[/edit]