Let $\alpha : M \rightarrow N$ and $\beta : L \rightarrow N$ be morphisms and $f: P\ \rightarrow L$ and $g:P \rightarrow M$ be the pullback of $(\alpha,\beta)$. If $\alpha$ is an epimorphism then $\ker (\alpha)$ is a direct summand of $P$ if and only if there exists $\kappa : L \rightarrow M$ such that $\beta=\alpha\kappa$.
I have allready proved that if the $\ker(\alpha)$ is a direct summand of $P$ then such $\kappa $ exists but I seem to get stuck trying to prove the other implication.
We see $\ker\alpha$ inside $P=\left\{(m,n):\alpha(m)=\beta(n)\right\}$ as $\ker\alpha\times 0$.
It actually is easier to start with the other direction, so we have an idea of how $\kappa$ should be constructed, so I believe you should look at your proof again:
If such $\kappa$ exists, then you should prove that $P=\left\{(m+\kappa(l),l):(m,l)\in \ker\alpha\times L\right\}=(\ker\alpha\times 0)\oplus(\kappa,1)(P)$. I'll leave the work for you.
This description of $P$ gives us some direction to construct $\kappa$: Assume that $\alpha$ is surjective and that $\ker\alpha$ is a (inner) direct summand of $P$: $P=(\ker\alpha\times 0)\oplus R$ for some submodule $R$. Given $n\in N$, there exists $m$ for which $\alpha(m)=\beta(n)$, so that $(m,n)\in P$. We can then write $(m,n)=(k,0)+(x,y)$ for a unique choice of $k\in\ker\alpha$ and $(x,y)\in R$. Of course then $y=n$. The formula given above (in the case that we already know $\kappa$) says that $\kappa(n)$ should be defined as $\kappa(n)=x$. So let's show that the choice of $m$ does not interfere with the value of $x$: Suppose we had chosen another $m'$ with $\alpha(m')=\beta(n)$, and again take the unique $k'\in \ker\alpha,x'\in M$ with $(m',n)=(k',0)+(x',n)$. Then $\alpha(x)=\alpha(m)=\alpha(m')=\alpha(x')$ and so $(x'-x,0)=(x',y)-(x,y)\in(\ker\alpha\times 0)\cap R=0$, hence $x'=x$. Thus $\kappa(n)=x$ is well-defined, and clearly $$\alpha(\kappa(n))=\alpha(x)=\alpha(m-k)=\alpha(m)=\beta(n)$$ because $k\in\ker\alpha$.