TL;DR: This can be seen by decoding the $k$ logical qubits out of the code subspace into the first $k$ physical qubits while keeping track of the stabilizer group.
Code subspace
Consider an $[\![n,k]\!]$ quantum error correcting code. Let $\mathcal{H}$ denote the $2^n$-dimensional Hilbert space of the block of $n$ physical qubits and $\mathcal C\subset\mathcal H$ the $2^k$-dimensional code subspace of $k$ logical qubits. Note that $\mathcal H$ contains $2^{n-k}$ mutually orthogonal $2^k$-dimensional subspaces isomorphic to $\mathcal C$. We can index those subspaces with binary strings $s=s_1s_2\dots s_{n-k}$ where $s_i\in\{0,1\}$. Let $\mathcal C^s$ denote the subspace corresponding to the binary string $s$. By convention $\mathcal C^0=\mathcal C$. The binary strings $s$ are called syndromes due to their role in error diagnostics.
Decoding unitary
Let $U$ denote the decoding unitary acting on a block of $n$ qubits. Suppose that $U$ sends the encoded state $|\psi\rangle$ into qubits $1\dots k$ and the syndrome $s$ into qubits $k+1\dots n$. More precisely, suppose that
$$
U|\tilde\psi^s\rangle=|\psi\rangle\otimes|s\rangle\tag1
$$
for $|\tilde\psi^s\rangle\in\mathcal C^s$. In particular, $U$ sends every logical state $|\tilde\psi\rangle\in\mathcal C$ to $|\psi\rangle\otimes|0_{n-k}\rangle$.
Decode logical subspace out into standalone qubits
Let $\mathcal C'$ denote the subspace of $\mathcal H$ consisting of states of the form $|\psi\rangle\otimes|0_{n-k}\rangle$. The stabilizer group $S'$ of $\mathcal C'$ consists of all $2^{n-k}$ operators of the form
$$
I_1\otimes I_2\otimes\dots\otimes I_k\otimes Z_{k+1}^{b_1}\otimes\dots\otimes Z_n^{b_{n-k}}\tag2
$$
where $b_i\in\{0,1\}$. Moreover, the normalizer $N(S')$ of $S'$ in $\mathcal G_n$ is equal to its centralizer in $\mathcal G_n$ and hence consists of all $4^{k+1}\cdot 2^{n-k}$ operators of the form
$$
i^fP_1\otimes P_2\otimes\dots\otimes P_k\otimes Z_{k+1}^{b_1}\otimes\dots\otimes Z_n^{b_{n-k}}\tag3
$$
where $f\in\{0,1,2,3\}$, $P_i\in\{I,X,Y,Z\}$ and $b_i\in\{0,1\}$. Comparing $(2)$ and $(3)$ we see that $N(S')\simeq\mathcal G_k\times S'$. Consequently, $N(S')/S'\simeq \mathcal G_k$.
Encode standalone qubits back into code subspace
Now, if $P$ stabilizes $|\psi\rangle\otimes|0_{n-k}\rangle=U|\tilde\psi\rangle$ then $U^\dagger PU$ stabilizes $|\tilde\psi\rangle$. Therefore, $U^\dagger S' U$ stabilizes $\mathcal C$ and thus $U^\dagger S' U=S$. However, conjugation by $U^\dagger$ is a group isomorphism, so $S\simeq S'$. Similarly, if $P$ and $Q$ commute (anti-commute), then $U^\dagger PU$ and $U^\dagger QU$ commute (anti-commute). Therefore, if $U$ is Clifford$^1$, then $N(S)\simeq N(S')$. Putting it all together we obtain
$$
N(S)/S\simeq N(S')/S'\simeq\mathcal G_k\tag4
$$
as expected.
$^1$ There is a subtlety here: the normalizer $N(S)$ of $S$ is defined with respect to the group containing $S$. In particular, the normalizer $N_{\mathcal G_n}(S)$ of $S$ in $\mathcal G_n$ is generally different from its normalizer $N_{U(2^n)}(S)$ in $U(2^n)$. In our case, we work with $N_{\mathcal G_n}(S)$, so the proof requires that $U$ be Clifford. This is in fact typically the case since we generally choose the logical Pauli operators from $\mathcal G_n$. In the more general case, the proof still works (since conjugation by any unitary, Clifford or otherwise, is a group isomorphism), but rigorous notation gets more complicated since we have to keep track of the subgroup of $U(2^n)$ that $\mathcal G_n$ gets mapped to.
