How to prove surjectivity part of Short Five Lemma for short exact sequences.

It should be straightforward.

We want to prove $\beta$ is surjective, so start out from an arbitrary element $b'\in B'$. We can do one thing: consider $c':=\phi'(b')\in C'$.
Since $\gamma$ is surjective, we get $c$ with $\gamma(c)=c'$.

That the pair $\phi,0$ of maps is exact means nothing else but that $\phi$ is surjective. It yields an element $b\in B$, such that $\gamma\phi(b)=c'=\phi'(b')$.

Now we might not get $\beta(b)=b'$ with this element $b$, nevertheless, we have that $\beta(b)$ and $b'$ has the same image under $\phi'$, so by exactness, this gives $a'\in A'$ such that $\psi'(a')=b' - \beta(b)$.

Can you take it on from here?

Thanks to Berci and Pedro Tamaroff.

Let $b' \in B'$. Then $\phi'(b') = c' \in C'$ and by surjectivity of $\gamma$, there's $c \in C$ with $\gamma(c) = \phi'(b')$ and there's $b \in B$ with $\phi(b) = c$. So $\gamma\phi(b) = \phi'(b') = \phi'(\beta(b))$. So $\beta(b) - b' \in \ker \phi' = \psi'(A') = \psi' \alpha(A)$. So there's $a \in A$ so that $\psi'\alpha(a) = \beta(b) - b' = \beta \psi(a)$. Then clearly $b'$ can be written as the image of the element $b - \psi(a)$ under $\beta$. We're done.