$X$ be the space obtained from attaching $D^{n+1}$ to $S^n$ by degree $m$ map. The cellular chain complex of the space is the following
$$0 \to \Bbb Z \stackrel{\times m}{\to} \Bbb Z \to 0 \to \cdots \to 0 \to \Bbb Z \to 0$$
since we have cells only at dimension $0, n$ and $n+1$, with a single cell in each of these dimensions. The chain complex has nontrivial homology only at the $n$-th level, where $H_{n}(X) \cong \Bbb Z/m\Bbb Z$. Thus, the space is an $M(\Bbb Z/m, n)$-space.
If you don't know cellular homology, use the long exact sequence for $(X, S^n)$ :$$0 \cong H_{n+1}(S^n) \to H_{n+1}(X) \to H_{n+1}(X, S^n) \stackrel{\partial}{\to} H_n(S^n) \to H_n(X) \to H_n(X, S^n) \cong 0$$
Note $H_{n+1}(X, S^n) \cong H_{n+1}(X/S^n) \cong H_{n+1}(S^{n+1}) \cong \Bbb Z$ (as $(X, S^n)$ is a CW-pair) and $H_n(S^n) \cong \Bbb Z$.
Recall that $\partial : H_{n+1}(X, A) \to H_n(A)$ sends homology class of $(n+1)$-cycles $\xi$ in $X$ relative to $A$ to homology class of the $n$-cycle $\partial \xi$ in $A$. [Hatcher, pg. $117$] Bearing that interpretation in mind, note that the generator of $H_{n+1}(X, S^n)$ corresponding to $1 \in \Bbb Z$ is represented in simplicial homology as the homology class of the relative cycle $\zeta$ in $Z^{n+1}(X, S^n)$ corresponding to a triangulation of the $(n+1)$-disk in $X$ by $(n+1)$-simplices with the sum of the faces being a cycle in $Z^n(S^n)$. This cycle in $Z^n(S^n)$ is precisely $\partial \zeta$, which is in turn represented by the degree $m$ map $S^n \to S^n$ in singular homology, as $X = D^{n+1} \cup_\varphi S^n$ where $\varphi$ is a degree $m$ map.
Thus, we conclude that $\partial$ sends $1$ to $m$. As $\partial$ is injective, $H_{n+1}(X) \cong \text{ker} \partial \cong 0$. The above sequence then boils down to the short exact sequence
$$0 \to \Bbb Z \stackrel{\times m}{\to} \Bbb Z \to H_n(X) \to 0$$
Hence, $H_n(X) \cong \Bbb Z/m\Bbb Z$. The same argument can be used to prove that $H_i(X) \cong 0$ for other $i$'s. Thus, we conclue $X$ is a $M(\Bbb Z/m\Bbb Z, n)$-space.
Someone suggested that I should provide a rigorous algebraic argument for the claim that $\partial$ is the multiplication by $m$ morphism. So here goes:
$\psi : D^{n+1} \hookrightarrow D^{n+1} \coprod S^n \to X$ be the map obtained from composing the inclusion into the disjoint union with the quotient map. $\psi|_{\partial D^{n+1}} : S^n \to S^n$ is clearly a degree $m$ map. We thus have a map of pairs $\psi : (D^{n+1}, S^n) \to (X, S^n)$. Use long exact sequence of both pairs coupled with naturality to obtain the commutative diagram
$$\require{AMScd} \begin{CD} H_{n+1}(D^{n+1}, S^n) @>>> H_n(S^n)\\ @VVV @VVV \\ H_{n+1}(X, S^n) @>>> H_n(S^n) \end{CD}$$
Note that the map $H_{n+1}(\psi) : H_{n+1}(D^{n+1}, S^n) \to H_{n+1}(X, S^n)$ can be canonically identified with the map $H_{n+1}(\bar{\psi}) : H_{n+1}(S^{n+1}) \to H_{n+1}(S^{n+1})$ where $\bar{\psi} : S^{n+1} \to S^{n+1}$ is induced from $\psi$ by quotienting the subspace $S^n$ in both spaces. One can easily see by the local degree formula that $\bar{\psi}$ is a map of degree $1$, hence $H_{n+1}(\bar{\psi})$ and thus $H_{n+1}(\psi)$ is ultimately an isomorphism. The top horizontal map is an isomorphism. The right vertical map is multiplication by $m$ because $\psi|_{\partial D^{n+1}}$ is of degree $m$. By commutativity of the diagram, the bottom horizontal map - the snake map $\partial$ - has to be multiplication by $m$ too. $\blacksquare$
Best Answer
Let $M_\gamma$ be the mapping cylinder of $\gamma$, which is homotopy-equivalent to $M(G, n)$. Then the mapping cone $C_\gamma$ sits in a cofibre sequence
$$ M(K, n) \to M_\gamma \to C_\gamma $$
and therefore we get a long exact sequence of homology groups. This sequence is mostly zero, except the following segment:
$$ \dots \to 0 \to H_{n+1}(C_\gamma) \to H_n(M(K, n)) \to H_n(M_\gamma) \to H_n(C_\gamma) \to 0 \to \dots$$
Then $H_n(M(K, n)) \cong K$ and $H_n(M_\gamma)\cong H_n(M(G, n)) \cong G$ and the homomorphism is given by inclusion and hence is injective so $H_{n+1}(C_\gamma) \cong 0$, and by the first isomorphism theorem $H_n(C_\gamma) \cong G/K$.