**Proof.**
If $M$ is faithfully flat, then $0 \to \mathop{\mathrm{Ker}}(\alpha ) \to N \to N'$ is exact if and only if the same holds after tensoring with $M$. This proves (1) implies (2). For the other, assume (2). Let $N_1 \to N_2 \to N_3$ be a complex, and assume the complex $N_1 \otimes _ R M \to N_2 \otimes _ R M \to N_3\otimes _ R M$ is exact. Take $x \in \mathop{\mathrm{Ker}}(N_2 \to N_3)$, and consider the map $\alpha : R \to N_2/\mathop{\mathrm{Im}}(N_1)$, $r \mapsto rx + \mathop{\mathrm{Im}}(N_1)$. By the exactness of the complex $-\otimes _ R M$ we see that $\alpha \otimes \text{id}_ M$ is zero. By assumption we get that $\alpha $ is zero. Hence $x $ is in the image of $N_1 \to N_2$.
$\square$

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