If you are happy with Frechet bundles here is an alternative approach. Let $\pi \colon {\mathcal G} \times M \to M$ be the projection and consider $\pi^{-1}(TM) \to {\mathcal G} \times M$ a real vector bundle of rank $n$. Following the notation in the paper let $P_{GL^+} \to M$ be the $GL^+(n, {\mathbb R})$ bundle of oriented frames of $TM$. The bundle of oriented frames of $\pi^{-1}(TM)$ is $\pi^{-1}(P_{GL^+})$. As in the paper pick a lift $P_{\widetilde{GL}^+} \to M$ of $P_{GL^+}$ to $\widetilde{GL}^+(n, {\mathbb R}) \to M$ of bundles over $M$. This exists because we assume $M$ is spin. Then $\pi^{-1}(P_{\widetilde{GL}^+})$ is a lift of $\pi^{-1}(P_{GL^+})$ to $\widetilde{GL}^+(n, {\mathbb R})$.
If $g \in {\mathcal G}$ and $m \in M$ then $\pi^{-1}(TM)_{(g, m)} = T_m M$ so has on it an inner product defined by $g(m)$. Denote this "universal" inner product on $\pi^{-1}(TM)$ by $g$. It will be smooth for the usual reason with Frechet manifolds which is because if $M$ and $N$ are finite-dimensional bundles then the evaluation map $$ M \times C^\infty(M, N) \to N $$ is a smooth map of Frechet manifolds [1]. Again following the approach in the paper we let $P_{SO} \subset \pi^{-1}(P_{GL^+}) $ be the subbundle of oriented orthonormal frames for the metric $g$. Taking the pre-image of this in $\pi^{-1}(P_{\widetilde{GL}^+})$ gives us a $Spin(r, s)$ bundle over $\mathcal{G} \times M$. The associated vector bundle to this using the spin representation gives us $E$ as a smooth, finite rank, Frechet vector bundle.
Finally you want a theorem that says that when you "push-down" $E$ with $\pi$ the result is a smooth Frechet vector bundle on ${\mathcal G}$. This seems reasonably but I'm not sure where to find it. I can't see it in [1].
Sorry this is a bit sketchy but that reflects the sketchiness of my knowledge of Frechet manifolds.
[1] Richard Hamilton -- The Inverse Function Theorem of Nash Moser. http://dx.doi.org/10.1090%2FS0273-0979-1982-15004-2
