 Solution

Write the two arrow matrices as $(A,B)\in M_2(k{)}^2$. The base change action on quiver representations sends them to $(g_{2}A{g}_1^{-1},g_{3}B{g}_2^{-1})$. Starting from $(I,I)$ yields precisely the pairs with both matrices invertible: given such a pair, choose $g_1=I$, $g_2=A$, $g_3=BA$. HenceThis is a nonempty Zariski-open subset of the irreducible affine space $M_2(k{)}^2$, so its closure is the entire representation space. Its boundary in that closure is $\{\det A\det B=0\}$.

The rank classification of a two-step linear map says an orbit is determined by $(r,s,t)=(\mathrm{rank}A,\mathrm{rank}B,\mathrm{rank}(BA))$. Indeed, the six interval multiplicities from the elementary decomposition areThey are nonnegative exactly when $0\le r,s\le 2$ and $\max (0,r+s-2)\le t\le \min (r,s)$. Apart from the open orbit $(2,2,2)$, there are nine boundary orbits. Put $D=\mathrm{diag}(1,0)$ and $E=\mathrm{diag}(0,1)$; representatives areThe two rank-one/rank-one cases differ by whether $\mathrm{im}A\subseteq \ker B$; the individual arrow ranks alone do not distinguish them.