Consider an exact sequence $$0\to M'\to M\to M''\to 0$$ of $R$-modules. Set $\iota:M\to M,\pi:M\to M''$. Let $P\in \operatorname{Ass}(M)-\operatorname{Ass}(M')$ be a prime ideal annihilating $m_0\in M$ and consider the injection of $R$-modules $\phi: R/P\to M, \bar r\mapsto rm_0$ induced by the map $R\to M, r\mapsto rm_0$.
First, for any nonzero element $rm$ in the image of $\phi$, $\operatorname{Ann}(rm)=P$. How does this imply $\operatorname{im}\phi\cap M'=0$? Suppose by contradiction that there is a nonzero $x=rm_1\in \operatorname{im}\phi\cap M'$. We need to prove that there is an injective map of $R$-modules $\psi: R/P\to M'$ (this will contradict that $P\notin \operatorname{Ass}(M')$). Consider the map $R\to M',r\mapsto rm_1$. We need to show that this map factors through $R/P$, for which we need to show that every element of $P$ maps to zero. But we only know $\operatorname{Ann}(rm_1)=P$ (since $rm_1\in \operatorname{im}\phi$). This doesn't show that $xm_1=0$ for all $x\in P$...
Second, why does $\ker (\pi\circ \phi)\simeq \operatorname{im}\phi\cap M'$? I don't accept the version that says "we can view $M'$ as a submodule of $M$" (because I don't find it rigorous enough). Officially, using exactness of the sequence, $\ker (\pi\circ \phi)=\{x\in R/P:\phi(x)\in \operatorname{im}(\iota) \}$. I don't see why this is $im\phi\cap M'$.