Say the maps in your first displayed diagram are, left to right between the first two rows, $\psi, f_1, f_0$ and between the second two rows $\phi, g_1, g_0$. Let $\sigma \partial_2 = \phi-\psi$ (I'm afraid my maps compose in the opposite direction to yours). Let the map $N \to Y_1$ be $\iota_Y$. You ask about the maps in the displayed equation following "Benson simply writes the diagram". Let's replace the middle $N$ by $P_2/\ker \phi$: then the maps between the first two rows are $\bar{\phi}$, $$p_1+\partial_2\ker\phi \mapsto f_1p_1 + \iota_Y \sigma p_1$$ (this turns out to be well-defined), $f_0$ then the identity. By $\bar{\phi}$ I mean the isomorphism induced by $\phi$ from $P_2/\ker\phi \to N$. Between the second two rows they are $\bar{\phi}, \bar{g_1}, g_0$, and the identity. Note that we don't need to assume $\ker \phi = \ker \psi$ or that $P_1/\partial_2\ker\phi = P_1/\partial_2 \ker \psi$.
Say two exact sequences beginning with $N$ and ending with $M$ are linked if there is a chain map between them acting as the identity on $M$ and $N$. Two such sequences represent the same element of $\operatorname{Ext}^2$ if and only if they are related by the equivalence relation generated by linkage. They need not be linked themselves, unlike for $\operatorname{Ext}^1$. So I am not sure a map between the last two sequences you write down will exist in general.
