I don't understand how from last step of the proof that $0 \rightarrow \operatorname{Hom}(\Omega_{C/B},N)\rightarrow \operatorname{Hom}(\Omega_{C/A},N)\rightarrow \operatorname{Hom}(\Omega_{B/A}\otimes C,N)$ the proposition follows? Why does that imply exactness?
The lemma they're using without making it so explicit is the following:
$X\stackrel{f}\to Y\stackrel{g}\to Z$ is exact for $R$-modules $X,Y,Z$ iff $ \hom_R(Z,M)\to \hom_R(Y,M) \to \hom_R(X,M)$ for $M$ an arbitrary $R$-module.
(They're actually using this twice: once for $ \Omega_{B/A}\otimes_BC\to\Omega_{C/A}\to \Omega_{C/B}$ and once for $\Omega_{C/A}\to\Omega_{C/B}\to0$.)
The forward direction is clear from the fact that $\hom$ is left-exact (it's a right adjoint, or you can check this directly), so what you're really after here is proving the reverse direction. Here's a proof:
Pick $M=Z$. Follow $id_Z$ to see that it is equal to both $g\circ f$ (as the maps between homs are the appropriate compositions) and $0$ (by exactness) in $\hom_R(X,Z)$ which shows that $\mathrm{im} f\subset \ker g$.
To show that $\mathrm{im} f= \ker g$ , let $M=Y/\mathrm{im} f$ and track the quotient map $q:Y\to Y/\mathrm{im}f$. By exactness of the hom sequence, there is $h:Z\to M$ such that $h\circ g=q$. But this means that the kernel of $g$ can't be larger than the image of $f$, as otherwise $q$ could not factor through it. So the sequence is exact.