In Atiyah and MacDonald, Prop 2.18 establishes that for any exact sequence $$M'\xrightarrow{f}M\xrightarrow{g}M''\xrightarrow{}0\tag{1}$$ of $A$-modules and homomorphisms, and for any $A$-module $N$, $$M'\otimes N\xrightarrow{f\otimes 1}M\otimes N\xrightarrow{g\otimes 1}M''\otimes N\xrightarrow{}0\tag{2}$$ is exact, where $1$ is the identity map. However, I'm a bit confused about a technique that they use in the proof. $E$ is defined to be the sequence (1), $E\otimes N$ is defined to be the sequence (2), and for any $A$-module P, they state that $\mathrm{Hom}(E,\mathrm{Hom}(N,P))$ must be exact, and therefore $\mathrm{Hom}(E\otimes N, P)$ is exact, and therefore $E\otimes N$ must be exact.
This seems to be referencing two earlier lemmas. The first, is that for $A$-modules $M$, $N$, $P$, we have $\mathrm{Hom}(M\otimes N, P)\cong \mathrm{Hom}(M,\mathrm{Hom}(N,P))$ by a canonical isomorphism. The second is that for any sequence $$M'\xrightarrow{u}M\xrightarrow{v}M''\xrightarrow{}0\tag{3}$$ of $A$-modules and homomorphisms, letting $N$ be an $A$-module, and defining $$0\xrightarrow{}\mathrm{Hom}(M'',N)\xrightarrow{\bar{v}}\mathrm{Hom}(M,N)\xrightarrow{\bar{u}}\mathrm{Hom}(M',N)\tag{4}$$ with $\bar{v}(\phi)=\phi\circ v$ and $\bar{u}(\phi)=\phi\circ u$, then (3) is exact if and only if (4) is exact $\forall\, N$.
I'm comfortable with both of these lemmas, but I'm struggling to understand the generalization made in the proof of the proposition. What exactly is meant by $E$ as a mathematical object, and why does it make sense to consider $\mathrm{Hom}(E,\mathrm{Hom}(N,P))$?
$E$ can be thought of as a complex (in this case a very special complex). For example, $E$ could stand for any complex of $A$-modules, $\{M_n\}, n\in\mathbb{Z}$ with module maps, say $\partial_n:M_n\to M_{n-1}$ for all $n$. It is a complex just means $\partial_{n-1}\circ\partial_n=0$ for all $n$. For any other module $P$, one has $\mathrm{Hom}(E,P)$ which is just the complex got by $\{\mathrm{Hom}(M_n,P)\}$ and the naturally induced maps.