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 $H_n(C_\gamma) \cong G/K$.
I'm going to ignore your approach to suggest one which gives you the maximal compact quickly. It won't use much different than your ideas of playing with fiber sequences.
Consider the space $\mathcal S_{p,q}$ of splittings $\Bbb R^{p,q}$ as $V \oplus W$, where $V$ is a $p$-dimensional subspace where the metric is positive definite and $W$ is a $q$-dimensional subspace on which the metric is negative definite. Topologize this as a subspace of the quotient $GL(p+q)/GL(p) \times GL(q)$.
Observe that $SO^+(p,q)$ acts properly and transitively on $\mathcal S_{p,q}$. (I will leave this part of the argument to you, it is straightforward.)
Less obvious is that $\mathcal S_{p,q}$ is contractible. This will use a fiber sequence argument.
Write $\mathcal V_{p,q}$ for the space whose elements are bases $\{x_1, \cdots, x_p, y_1, \cdots, y_q\}$ for $\Bbb R^{p,q}$, where the metric is positive definite on the span $\langle x_1, \cdots, x_p\rangle$ and negative definite on the span $\langle y_1, \cdots, y_q\rangle$, topologized as a subspace of $GL(p+q)$. Then we have a fiber sequence $$GL(p) \times GL(q) \to \mathcal V_{p,q} \to \mathcal S_{p,q}.\require{AMScd}$$
We also have a fiber sequence $\mathcal V_{p-1,q} \to \mathcal V_{p,q} \to (\Bbb R^p \setminus 0)$, compatible with the actions of $GL(p-1) \times GL(q)$, which descends to a map $\mathcal S_{p-1, q} \to \mathcal S_{p,q}$; send $(V,W) \mapsto (V \oplus \langle e_p\rangle, W)$, where $e_p$ is the last basis vector in $\Bbb R^p$. So we have obtained the diagram
$$\begin{CD}
GL(p-1) \times GL(q) @>>> GL(p) \times GL(q) @>>> (\Bbb R^p \setminus 0) \\
@VVV @VVV @|\\
\mathcal V_{p-1,q} @>>> \mathcal V_{p,q} @>>> (\Bbb R^p \setminus 0)\\
@VVV @VVV\\
\mathcal S_{p-1,q} @>>> \mathcal S_{p,q}
\end{CD}$$
Now induct. If $\mathcal S_{p-1,q}$ is contractible, then the top-left vertical arrow is an equivalence, so by the 5-lemma the middle vertical arrow is an equivalence, and hence $\mathcal S_{p,q}$ is contractible. A nearly identical argument allows you to reduce the value of $q$.
For the base case, $\mathcal S_{0,0}$ is the one-point space. Therefore $\mathcal S_{p,q}$ is contractible for all $(p,q)$.
Because the stabilizer of $(\Bbb R^p, \Bbb R^q)$ is $SO(p) \times SO(q)$, we find we have a fiber sequence $$SO(p) \times SO(q) \to SO^+(p,q) \to \mathcal S_{p,q}$$ with contractible base, so that the inclusion $SO(p) \times SO(q) \to SO^+(p,q)$ is an equivalence.
I suspect there is a non-inductive, geometric argument that $\mathcal S_{p,q}$ is contractible using that the subspaces must intersect the null-cone trivially.
Best Answer
To elaborate on my comment:
Yes, $1$-connected means simply connected.
$H_i(X)$ means $H_i(X, \mathbb{Z})$ by definition.
Both definitions have small errors in them: in Arkowitz's definition he should either be talking about reduced homology or otherwise excluding the case $i = 0$, and your professor's definition is missing the simply connected condition. Without that condition the definition is satisfied by homology spheres such as the Poincaré homology sphere, which are not simply connected and are not Moore spaces. However, genuine spheres $S^n$ are Moore spaces $M(\mathbb{Z}, n)$ for $n \ge 2$.