Here's a picture of the Snake Lemma from nLab:
I'm having trouble showing exactness at $\ker g$ and haven't even got to the connecting hom yet. So let's focus on $\ker g$. Let $q : A' \to B'$ be the hom such that $\text{im} \ q = \ker p$. Then for one direction, namely $\text{im} \ \overline{q} \subset \ker \overline{p}$ we have $x = \overline{q}(x') = q(x)$ since $\overline{q}$ is simply a restriction of $q$. But $p \circ q (x) = 0$ by assumption of exactness of rows of the central diagram. Thus $\overline{p} \circ \overline{q} (x) = 0$ since $\overline{q}, \overline{p}$ are simply restrictions of $q, p$. But that means precisely that $\text{im} \ q \subset \ker p$.
I'm having trouble showing the other direction, $\ker \overline{p} \subset \text{im} \ \overline{q}$. If $\overline{p}(x) = 0$, then $p(x) = 0 \implies x = q(x')$ for some $x' \in A'$. But how do I show then that $x' \in \ker f$ necessarily?
The trick is to use commutativity.
If $x\in \ker g$ with $p(x)=0$, then as you've noted, there is $x'\in A'$ with $q(x')=x$.
Then $$0=g(x)=gq(x')=if(x'),$$ but by exactness of the second row, $i$ is injective, so $f(x')=0$. Therefore $x'\in \ker f$, as desired.