Let $Q=(Q_0,Q_1,s,t)$ be an acylic connected finite quiver where $Q_0$ are vertices, $Q_1$ are arrows, and $s,t:Q_1\to Q_0$ are source and target respectively. Let $K$ be a field. Let $KQ$ be the associated path algebra.
"$Cok(f:\oplus_{x\in Q_0}Hom_k(M_x,M'_x)\to\oplus_{a\in Q_1}Hom_k(M_{s(a)},M'_{t(a)}))=Ext^1(M,M')$"
The map above $f:(f_x)_{x\in Q_0}\to (f_{s(a)}\alpha-\alpha f_{t(a)})_{a\in Q_1}$ where $f:M\to M'$ is given by a family of morphisms $f_x$ and $\alpha\in Q_1$ is treated as morphism for $M_{s(\alpha)}\to M_{t(\alpha)}$ and $M'_{s(\alpha)}\to M'_{t(\alpha)}$ .
$\textbf{Q:}$ How do I see the identification $Coker(f)=Ext^1(M,M')$?
Ref. Ringel, What is a hereditary algebra? Sec 3.
There is a standard resolution of $M$ of the form $$0 \rightarrow P_1 \rightarrow P_0 \rightarrow M \rightarrow 0.$$ Applying $\textrm{Hom}(\cdot,M')$, you get a long exact sequence: $$ 0\rightarrow \textrm{Hom}(M,M') \rightarrow \textrm{Hom}(P_0,M') \rightarrow \textrm{Hom}(P_1,M') \rightarrow \textrm{Ext}^1(M,M') \rightarrow \textrm{Ext}^1(P_0,M')$$
Now $\textrm{Ext}^1(P_0,M')=0$ since $P_0$ is projective, so you actually have a four-term exact sequence. The map between the second and third terms can be identified with the one that you ask about, so its cokernel is isomorphic to $\textrm{Ext}^1(M,M')$.
Section 5.6 and 5.7 of https://www.math.uni-bielefeld.de/~sek/kau/leit5.pdf give a lot of details. This resolution of $M$ is called the "standard resolution"; you should be able to find it discussed in plenty of places.