Show that $k[x]/(x^{2})$ is an indecomposable (1), but not irreducible (2) $k[x]$-module. Exercise: Show that $k[x]/(x^{2})$ is an indecomposable (1), but not irreducible (2) $k[x]$-module.

I'm not sure about all different kind of modules, but this is a question of a book about associative algebras. It is not really exercise, it is just stated in the text, so  I guess it must be rather trivial, but I'm not getting it. I think it may be I forgot some old ring theory/linear algebra stuff.
If I understand correctly: $k[x]/(x^{2})$ can be seen as $\{a+bx: a,b \in k\}$. 
So for (1), I must show that there don't exist two non-zero representation $V_{1},V_{2}$ of $k[x]$so that $k[x]/(x^{2})$ is isomorphic to $V_{1}\oplus V_{2}$. Well as $k[x]/(x^{2})$ is 2 dimensional. $V_{1},V_{2}$ have to be $1$ dimensional. But I'm not sure how the 1-dimensional subrepresentation look like. Are they just isomorphic to $k$ ? 
For (2), I need to show there exist an non-trivial subrepresention. So I have to show there exists a subspace $W$ of $\{a+bx: a,b \in k\}$ such that $fW \subset W$ for any polynomial $f$. Well, that seems impossible to me. I can't think of any if $W=k$, then $x\cdot \alpha$ is not in $W$. If $W=\{bx:b\in k\}$ then $bx bx$ is not in $W$.

I feel like I look at this the complete wrong way, feel free to ignore all my stuff above, and show me a way how to look at this. :)

For a field $k$, $k[x]$ is a principal ideal ring. By the correspondence theorem, the only ideals of $k[x]/(x^2)$ are those generated by divisors of $x^2$. Thus $k[x]/(x^2)$ has exactly three submodules: $(x^2)/(x^2),(x)/(x^2)$ and $k[x]/(x^2)$.
So the existence of $(x)/(x^2)$ immediately tells you why $k[x]/(x^2)$ isn't irreducible. And how could it be directly decomposable? There is only one proper ideal, so there isn't a second proper ideal to complement the first. So, there isn't a nontrivial decomposition of $k[x]/(x^2)$
