For the quiver $1 \rightarrow 2$, we can consider the category of representations $Rep \;(Q)$. The objects are representations of this quiver, and the morphisms are such that we have
$$\require{AMScd}f_1 \begin{CD} V_1 @>{\phi}>> V_2\\ @VVV \circlearrowright @VVV \\ V_1' @>{\phi'}>> V_2'; \end{CD}f_2$$
(i.e. the diagram commutes).
I am told that this is category is fully decomposable, but I don't know what that means. I'm guessing that every object in this category can be somehow written as a sum of other indecomposable objects, but I'm not clear about in what sense I mean the word "sum", here. Clarification would be appreciated.
If we're talking representations over a field, or more generally over a PID, then every representation is a direct sum of irreducible ones, as in the representation theory of finite groups over fields of characteristic zero. The irreducible representations over a field are represented by the multiplication maps $m_y:x\mapsto xy$ for each nonzero $y\in R$, as well as by $0\to R$ and $R\to 0$. $m_y$ and $m_z$ are isomorphic if and only if $y=uz$ for some unit $u$, so there are three irreducibles in total. Over a PID, there are more irreducibles, since there are nonunits and also nonfree modules.
The direct sum in your category is given by the diagonal sum: if $f:A\to B,g:C\to D$, then there is an induced map $f\oplus g:A\oplus C\to B\oplus D$ sending $(a,c)$ to $(fa,gc)$. Then the proof of "complete decomposability" for this category follows from the Smith normal form, which shows exactly that any map of (free) modules can be diagonalized, which decomposes it into irreducibles maps.