For now let's consider the $n=2$-subspace quiver: $Q = \bullet \to \bullet \leftarrow \bullet$. I have shown (with the help of Kirillov) that to classify these representations, it suffices to classify triples $V_0,V_1,V_2$ with $V_1,V_2$ subspaces of $V_0$. From linear algebra, given such a representation, we may always consider a basis of $V_0$ which consists of a basis of $V_1\cap V_2,V_1$ and $V_2$. This is all understood to me. Then Kirillov states:
"In terms of quiver representations, this can be rewritten as follows: any representation of the quiver $Q$ is isomorphic to a direct sum of the following representations:" $$ \overset{k}{\bullet}\to \overset{0}{\bullet} \leftarrow \overset 0 \bullet,\quad \quad \overset 0 \bullet \to \overset 0\bullet \leftarrow \overset k\bullet,\quad \quad\overset 0 \bullet \to \overset k \bullet \leftarrow \overset 0 \bullet $$ $$ \overset{k}{\bullet}\to \overset{k}{\bullet} \leftarrow \overset 0 \bullet,\quad \quad \overset 0 \bullet \to \overset k\bullet \leftarrow \overset k\bullet,\quad \quad\overset k \bullet \to \overset k \bullet \leftarrow \overset k \bullet $$ Notably it seems like every possible combination of either putting $k$ or $0$ at every vertex, excluding the possibility of $k$ on both outside vertices and $0$ in the middle, and excluding all $0$. I don't understand how to convert the linear algebra statement about equipping $V_0$ with a suitable basis to decomposing the given representation into this direct sum. Can anyone help explain this passage?
Consider a representation $\alpha : V_0 \longrightarrow V_1 \longleftarrow V_2: \beta$. If $W = \ker \alpha$ and if we choose a complement $V_0'$, then the representation decomposes into $(W,0,0)$ and $(V_0',V_1,V_2)$, and we can assume that $\alpha$ is injective. Similarly, we can assume that $\beta$ is injective. This gives $\dim\ker\alpha$ summands of the form $\mathbb k \to 0\leftarrow 0$ and $\dim\ker\beta$ summands of the form $0\to 0\leftarrow \mathbb k$.
The images of $\alpha$ and $\beta$ are now subspaces $A$ and $B$ of $V_1$, and they span a subspace $U$ of $V_1$. Let $U'$ be a complement of $U$ in $V_1$. Then there are $\dim U'$ copies of $0\to \mathbb k\leftarrow 0$ in your representation, and we can assume that $A+B= V_1$ and that we have inclusions $A\to V_1\leftarrow B$.
Now write $A\cap B = C$, then write $A = A'\oplus C$, $B = B'\oplus C$. Then you have $\dim C$ copies of $\mathbb k\to \mathbb k\leftarrow \mathbb k$ in your representation, and you can assume that $A\cap B = 0$. In this case, you get $\dim A$ copies of $\mathbb k\to \mathbb k\leftarrow 0$, and $\dim B$ copies of $0\to \mathbb k\leftarrow \mathbb k$. This finishes the classification.