Let $\Lambda$ be a cocommutative Hopf algebra over a commutative ring $R$. For two left $\Lambda$-modules $M$ and $N$, interpret $\mathrm{Ext}_{\Lambda}^n(M,N)$ as the set of equivalence classes of extensions of $M$ by $N$, which are exact sequences of left $\Lambda$-modules
$$0\to N\to A_{n-1}\to\ldots\to A_0\to M\to 0$$
with $n$ intermediate terms. On page $58$ of Benson's "Representations and Cohomology," Volume 1, there is a reference to a natural map
$$\mathrm{Ext}^n_{\Lambda}(R,R)\xrightarrow{\otimes_RM}\mathrm{Ext}_{\Lambda}^n(M,M)$$
given by "tensoring exact sequences with $M$." However, it is not clear to me that $-\otimes_RM$ is an exact functor, so if
$$0\to R\to A_{n-1}\to\ldots\to A_0\to R\to 0$$
is exact, why does it follow that
$$0\to M\to A_{n-1}\otimes_RM\to\ldots\to A_0\otimes_RM\to M\to 0$$
is also exact? Also, if I view $\mathrm{Ext}^n_{\Lambda}(R,R)$ as the set of equivalence classes of maps $\zeta:P_n\to R$ where $P^{\bullet}$ is a projective resolution of $R$, then how do I construct the image of the element represented by $\zeta$ under this natural map? (in terms of a map $Q_n\to M$ where $Q^{\bullet}$ is a projective resolution of $M$)
With regard to your first question: in that section, the assumption given at the start (that $M$ is a projective $R$-module) is in force throughout. This makes $ -\otimes_R M$ exact, since projective modules are flat. Otherwise you are right that there is no reason to think the sequence will remain exact after tensoring with $M$.
For your second question, the image of $\xi$ is the map $\xi\otimes \operatorname{id}_M : P_n \otimes M \to R \otimes M = M$. This makes sense because $P_* \otimes M$ is a projective resolution of $R\otimes M=M$, since (by 3.1.5 in your book) $P\otimes M$ is projective whenever $P$ is, and as above tensoring over $R$ with $M$ preserves exactness. If you have a fixed projective resolution $Q_*$ of $M$ you must compute a chain map $H_*: Q_* \to P_*\otimes M$ lifting the identity map $M \to M$ (see 2.4.2), and then the image of $\xi$ will be represented by $(\xi \otimes \operatorname{id}) \circ n : Q_n \to M$.