It is what you think it is:

**Lemma.** *The class of the short exact sequence $0 \to K \to M_n \to N \to 0$ in $\operatorname{Ext}^1_G(N,K)$ is $n$ times the class of the sequence $0 \to K \to M \to N \to 0$.*

*Proof.* By construction, the class in $\operatorname{Ext}^1_R(C,A)$ of a short exact sequence $0 \to A \to B \to C \to 0$ of $R$-modules is the image of $\operatorname{id}_C$ under the boundary map $\operatorname{Hom}(C,C) \to \operatorname{Ext}^1_R(C,A)$ coming from the long exact sequence
$$0 \to \operatorname{Hom}(-,A) \to \operatorname{Hom}(-,B) \to \operatorname{Hom}(-,C) \to \operatorname{Ext}^1_R(-,A) \to \ldots.$$
In our situation, we have a commutative diagram with exact rows
$$\begin{array}{ccccccccc}0 & \to & K & \to & M_n & \to & N & \to & 0 \\ & & |\!| & & \downarrow & & \downarrow{\scriptsize n}\!\!\!\! & & \\ 0 & \to & K & \to & M & \to & N & \to & 0.\!\end{array}$$
The long exact sequences give a commutative diagram
$$\begin{array}{ccccccccccc}0 & \to & \operatorname{Hom}(N,K) & \to & \operatorname{Hom}(N,M_n) & \to & \operatorname{Hom}(N,N) & \to & \operatorname{Ext}^1_G(N,K) & \to & \ldots \\ & & |\!| & & \downarrow & & \downarrow{\scriptsize n}\!\!\!\!\!\!\! & & |\!| & \\ 0 & \to & \operatorname{Hom}(N,K) & \to & \operatorname{Hom}(N,M) & \to & \operatorname{Hom}(N,N) & \to & \operatorname{Ext}^1_G(N,K) & \to & \ldots,\!\!\!\end{array}$$
hence the image of $\operatorname{id}_N$ in the first row maps to $n$ times the image of $\operatorname{id}_N$ in the second row. $\square$