The maps $\bar{f'} : \ker \alpha \to \ker \beta$ and $\bar{g'} : \ker \beta \to \ker \gamma$ are induced by restricting $f'$ and $g'$. If $\alpha(a') = 0$, then $\beta(f'(a')) = f(\alpha(a')) = 0$, so $f'(a') \in \ker \beta$. Similarly for $g'$.

The maps $\bar{f} : \mathrm{coker}\, \alpha \to \mathrm{coker}\, \beta$ and $\bar{g} : \mathrm{coker}\, \beta \to \mathrm{coker}\, \gamma$ are induced by $f$ and $g$. If $a \in \mathrm{im}\, \alpha$, say $a = \alpha(a')$, then $f(a) = f(\alpha(a')) = \beta(f'(a')) \in \mathrm{im}\, \beta$. So $f$ descends to $\bar{f} : A/\mathrm{im}\, \alpha \to B/\mathrm{im}\, \beta$.

We define $\delta : \ker \gamma \to \mathrm{coker}\, \alpha$.

Let $c' \in \ker \gamma$, so $\gamma(c') = 0$.

Lift: Since $g'$ is surjective, choose $b' \in B'$ with $g'(b') = c'$.

Map down: Compute $\beta(b') \in B$. We have $g(\beta(b')) = \gamma(g'(b')) = \gamma(c') = 0$. So $\beta(b') \in \ker g = \mathrm{im}\, f$ (by exactness of the bottom row at $B$).

Lift to $A$: Since $f$ is injective (bottom row is exact at $A$), there is a unique $a \in A$ with $f(a) = \beta(b')$.

Project: Define $\delta(c') = [a] \in \mathrm{coker}\, \alpha = A / \mathrm{im}\, \alpha$.

Well-definedness: Suppose $b_1'$ is another lift: $g'(b_1') = c'$. Then $g'(b' - b_1') = 0$, so $b' - b_1' \in \ker g' = \mathrm{im}\, f'$. Write $b' - b_1' = f'(a')$ for some $a' \in A'$. Then the corresponding elements $a, a_1 \in A$ differ by: $f(a - a_1) = \beta(b' - b_1') = \beta(f'(a')) = f(\alpha(a'))$. Since $f$ is injective, $a - a_1 = \alpha(a') \in \mathrm{im}\, \alpha$. So $[a] = [a_1]$ in $\mathrm{coker}\, \alpha$.

Homomorphism: $\delta$ is additive since each step (lift, map down, lift, project) is additive (using the uniqueness of the lift through $f$).

We show $\mathrm{im}(\bar{f'}) = \ker(\bar{g'})$.

$\mathrm{im}(\bar{f'}) \subseteq \ker(\bar{g'})$: For $a' \in \ker \alpha$, $\bar{g'}(\bar{f'}(a')) = g'(f'(a')) = 0$ by exactness of the top row.

$\ker(\bar{g'}) \subseteq \mathrm{im}(\bar{f'})$: Let $b' \in \ker \beta$ with $g'(b') = 0$. By exactness of the top row at $B'$, $b' = f'(a')$ for some $a' \in A'$. Then $f(\alpha(a')) = \beta(f'(a')) = \beta(b') = 0$. Since $f$ is injective, $\alpha(a') = 0$, so $a' \in \ker \alpha$ and $b' = \bar{f'}(a')$.

We show $\mathrm{im}(\bar{g'}) = \ker \delta$.

$\mathrm{im}(\bar{g'}) \subseteq \ker \delta$: Let $b' \in \ker \beta$, and set $c' = g'(b')$. To compute $\delta(c')$, we can use $b'$ itself as a lift. Then $\beta(b') = 0$, so $a = 0$ and $\delta(c') = [0] = 0$.

$\ker \delta \subseteq \mathrm{im}(\bar{g'})$: Let $c' \in \ker \gamma$ with $\delta(c') = 0$. Using a lift $b'$ with $g'(b') = c'$, we have $\beta(b') = f(a)$ and $[a] = 0$ in $\mathrm{coker}\, \alpha$, so $a = \alpha(a')$ for some $a'$. Then $\beta(b') = f(\alpha(a')) = \beta(f'(a'))$, so $\beta(b' - f'(a')) = 0$, meaning $b' - f'(a') \in \ker \beta$. And $g'(b' - f'(a')) = g'(b') - g'(f'(a')) = c' - 0 = c'$. So $c' = \bar{g'}(b' - f'(a'))$.

We show $\mathrm{im}\, \delta = \ker \bar{f}$.

$\mathrm{im}\, \delta \subseteq \ker \bar{f}$: Let $c' \in \ker \gamma$ with $\delta(c') = [a]$, where $f(a) = \beta(b')$ for a lift $b'$ of $c'$. Then $\bar{f}([a]) = [f(a)] = [\beta(b')] = 0$ in $\mathrm{coker}\, \beta$.

$\ker \bar{f} \subseteq \mathrm{im}\, \delta$: Let $[a] \in \mathrm{coker}\, \alpha$ with $\bar{f}([a]) = 0$, i.e., $f(a) \in \mathrm{im}\, \beta$. Write $f(a) = \beta(b')$ for some $b' \in B'$. Set $c' = g'(b')$. Then $\gamma(c') = \gamma(g'(b')) = g(\beta(b')) = g(f(a)) = 0$, so $c' \in \ker \gamma$. By construction, $\delta(c') = [a]$.

Similar to Step 3, using the dual argument. We omit the details. $\square$