I would definitely start by trying to convince myself the truth of the theorem using some heuristic reasoning. Among other things, I would mentally discard the short exact sequences in favor of submodules and cosets. In this case, we have a homomorphism $h\colon C\to C'$ between two modules with the following properties:
$h$ maps some submodule $A$ of $C$ isomorphically to a submodule $A'$ of $C'$.
Moreover, $h$ induces a bijection between the cosets of $A$ and the cosets of $A'$.
From this reasoning, it seems clear that $h$ will be a bijection, since the map from each coset of $A$ to the corresponding coset of $A'$ will resemble the isomorphism from $A$ to $A'$.
Next I would write out the diagram-chasing proof. However, I would be thinking in terms of submodules and cosets, and change the language to short exact sequences as I write. Here's the argument that would go through my head:
Injective: Suppose $h(c_1) = h(c_2)$ for some $c_1$ and $c_2$. Since $h$ acts bijectively on cosets, $c_1$ and $c_2$ must be in the same coset of $A$. Then $c_1 - c_2$ is an element of $A$ that maps to $0$ under $h$, so $c_1 = c_2$.
Surjective: Let $c' \in C'$. How can we find an element of $C$ that maps to it under $h$? Well, we know that $h$ acts bijectively on cosets, so we can find a coset $c+A$ that maps to $c'+A'$ under $H$. If we choose a representative $c$ of this coset, it won't necessarily be the case that $h(c) = c'$. However, it is true that $h(c) = c' + a'$ for some $a' \in A'$. Then $h(c-a) = c'$, where $a$ is the element of $A$ that maps to $a'$.
Here's what it would look like written out:
\begin{array}{ccccccc} 0 & \to & A & \xrightarrow{i} & C & \xrightarrow{\pi} & B & \to & 0 \\ & & \downarrow f & & \downarrow h & & \downarrow g \\ 0 & \to & A' & \xrightarrow{i'} & C' & \xrightarrow{\pi'} & B' & \to & 0 \end{array}
Injective: Let $c_1,c_2\in C$, and suppose that $h(c_1) = h(c_2)$. Then $$ g(\pi(c_1)) \;=\; \pi'(h(c_1)) \;=\; \pi'(h(c_2)) = g(\pi(c_2)). $$ Since $g$ is injective, it follows that $\pi(c_1) = \pi(c_2)$, so $\pi(c_1 - c_2) = 0$. It follows that $c_1 - c_2 = i(a)$ for some $a\in A$. Then $i'(f(a)) = h(i(a)) = h(c_1-c_2) = 0$. Since $i'$ and $f$ are injective, it follow that $a = 0$, so $c_1 = c_2$.
Surjective: Let $c'\in C'$. Since $g \circ \pi$ is surjective, there exists a $c\in C$ so that $g(\pi(b)) = \pi'(c')$. Then $\pi'(h(b)) = \pi' (c')$, so $h(b) - c' \in \ker(\pi')$. It follows that $h(c) - c' = i'(a')$ for some $a' \in A'$. Since $f$ is surjective, there exists an $a\in A$ so that $f(a) = a'$. Then $$ h(c-i(a)) \;=\; h(c) - h(i(a)) \;=\; h(c) - i'(f(a)) \;=\; h(c) - i'(a') \;=\; h(c) - (h(c) - c') \;=\; c'. $$