= Solution
Write the two arrow matrices as $(A,B)\in M_2(k)^2$. The <base change action on quiver representations> sends them to $(g_2Ag_1^{-1},g_3Bg_2^{-1})$. Starting from $(I,I)$ yields precisely the pairs with both matrices invertible: given such a pair, choose $g_1=I$, $g_2=A$, $g_3=BA$. Hence
$$
\boxed{\mathcal O_X=\{(A,B):\det A\det B\ne0\}\cong\operatorname{GL}_2(k)^2}.
$$
This 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 $(r,s,t)=(\operatorname{rank}A,\operatorname{rank}B,\operatorname{rank}(BA))$. Indeed, the six interval multiplicities from the elementary decomposition are
$$
m_{13}=t,\quad m_{12}=r-t,\quad m_{23}=s-t,\quad m_{11}=2-r,\quad m_{22}=2-r-s+t,\quad m_{33}=2-s.
$$
They are nonnegative exactly when $0\le r,s\le2$ and $\max(0,r+s-2)\le t\le\min(r,s)$. Apart from the open orbit $(2,2,2)$, \b[there are nine boundary orbits]. Put $D=\operatorname{diag}(1,0)$ and $E=\operatorname{diag}(0,1)$; representatives are
$$
\begin{array}{c|c|c}
(r,s,t)&A&B\\\hline
(0,0,0)&0&0\\
(0,1,0)&0&D\\
(0,2,0)&0&I\\
(1,0,0)&D&0\\
(2,0,0)&I&0\\
(1,1,0)&D&E\\
(1,1,1)&D&D\\
(1,2,1)&D&I\\
(2,1,1)&I&D
\end{array}
$$
The two rank-one/rank-one cases differ by whether $\operatorname{im}A\subseteq\ker B$; the individual arrow ranks alone do not distinguish them.
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