Ok I think I've got it, following the partition argument indicated by Michael Greinecker in the comments above. Thank you Michael! Please let me know if you spot any errors.
Lemma. Let $G$ be a locally compact group with Borel $\sigma$-algebra $\mathcal{B}(G)$. For each measurable function $k : G \to G$ and the push-forward measure $\kappa := k_* \eta = \eta \circ k^{-1}$ on $G$, there exists a disintegration $\eta_k(\mathrm{d} g|g')$ such that for any $B \in \mathcal{B}(G)$, \begin{equation} \eta(B) = \int_G \eta_k(B|g') \kappa(\mathrm{d} g') = \int_G \eta_k(B|k(g)) \eta(\mathrm{d} g). \end{equation}
Proof. We construct the global disintegration by a local partition argument. By local compactness of $G$, there exists a countable partition $\mathcal{C}$ of disjoint pre-compact sets of finite Haar measure, whose union equals $G$.
For each $C \in \mathcal{C}$, define the probability measure $\eta^C := \frac{1}{\eta(C)} 1_C \eta$. Each $\eta^C$ is Radon, so by the disintegration theorem (cf. Theorem 3.1 of Leao et al. 2004), there exists a regular conditional probability through $k$, i.e., a measurable measure-valued function $g' \mapsto \eta^C_k(\mathrm{d} g|g')$ such that the disintegration equation holds for each $B \in \mathcal{B}(G)$: \begin{equation} \frac{\eta(C \cap B)}{\eta(C)} = \eta^C(B) = \int_G \eta^C_k(B|g') \kappa(\mathrm{d} g') = \int_G \eta_k^C(B|k(g)) \eta(\mathrm{d} g). \end{equation}
We now define $\eta_k$ by combining across partition sets. Suppose $B \in \mathcal{B}(G)$ has finite Haar measure $\eta(B) < \infty$. Then: \begin{equation} \eta(B) = \sum_C \eta(C \cap B) = \int_G \sum_{C \in \mathcal{C}} \eta(C) \eta^C_k(B|g') \kappa(\mathrm{d} g') =: \int_G \eta_k(B|g') \kappa(\mathrm{d} g'), \end{equation} where we define $\eta_k(B|g') := \sum_{C \in \mathcal{C}} \eta(C) \eta^C_k(B|g')$.