Prop. 2.9 says the following:
A representation $Q$ is simple iff it is isomorphic to $S(i)$ for some $i \in Q_0$.
The proof given in the text starts off as follows:
It is clear that the $S(i)$ are simple representations. Conversely, let $M = (M_i, \varphi_\alpha)$ be any representation of $Q$. We want to show that there is a vertex $i$ such that $S(i)$ is a subrepresentation of $M$, and we have to choose this vertex $i$ carefully. We do not want to have a nonzero map in the representation $M$ that starts at the vertex $i$. For example, if $i$ is a sink in the quiver, we have what we want. But on the other hand, we also need the representation $M$ to be nonzero at the vertex $i$. This leads us to pick $i$ as follows.
Let $i \in Q_0$ such that $M_i \neq 0$ and $M_j = 0$, whenever there is an arrow $i \xrightarrow{\alpha} j$ in $Q$. Note that such a vertex exists since $Q$ has no oriented cycles. ...
I do not get exactly the reasoning for the choice of the vertex $i$. I know why it makes sense to require $M_i \neq 0$ and $M_j = 0$ so that it looks like $S(i)$ but why do we care that this be true whenever $\alpha$ exists? Also, how does having no oriented cycles guarantee that such a vertex $i$ exists?
As $S(i)$ is simple, any nonzero homomorphism $f:S(i)\to M$ is injective, and so its image will be a subrepresentation of $M$ isomorphic to $S(i)$. Thus to prove the proposition, we need to find some such nonzero map.
A homomorphism $f:S(i)\to M$ is given by a linear map $f_i:k\to M_i$ (where $k$ is the field we are working over) such that $\varphi_\alpha f_i=0$ for all arrows $\alpha$ starting at $i$. (Equivalently, we want an element $f_i(1)=m\in M_i$ in the kernel of all such $\varphi_\alpha$.)
As we don’t know much about the maps $\varphi_\alpha$, we would like to find some vertex $i$ such that $M_i\neq0$, but $M_j=0$ for all arrows $\alpha:i\to j$. Then necessarily $\varphi_\alpha=0$, and we are done.
This immediately happens if $i$ is a sink, since there are no arrows starting at $i$. If the quiver has no oriented cycles, then we can choose any vertex $p$ with $M_p\neq0$. If this does not satisfy our conditions, then there is an arrow $\alpha:p\to q$ with $M_q\neq0$. Repeating, we move along arrows, and since there are no oriented cycles, this process must terminate, say at vertex $i$. Then this $i$ satisfies our conditions.
N.B. $M$ will in general not be simple, so we cannot expect $M_j=0$ for all $j\neq i$, but it will contain a simple submodule. Take Schiffler’s Example 1.5 on pages 6,7. Then $M’$ is simple, and a submodule of $M’’$, but $M’’$ has total dimension 4.