I'm working over $\mathbb{C}$ here. If we let $G=\mathrm{SO}(2n+1)$ be the odd orthogonal group, and $P$ be the maximal parabolic corresponding to the $1$st node in the Type $B_n$ Dynkin diagram (following Bourbaki notation- I mean the endpoint which is adjacent to a doubled edge), then I believe $G/P$ is what is called the (maximal) orthogonal Grassmannian $\mathrm{OG}(n,2n+1)$.
The Borel-Weil theorem says that the $m$th homogeneous component of the coordinate ring of $G/P=\mathrm{OG}(n,2n+1)$ should be isomorphic to the irreducible representation $V^{m\omega_1}$, where $\omega_1$ is the corresponding fundamental weight. This should hold at least at say the level of representations of the Lie algebra $\mathfrak{g}=\mathfrak{so}(2n+1)$. Actually, it might be that it's the contragredient representation $(V^{m\omega_1})^*$ (because we're acting on functions), but I think that in Type B negation belongs to the Weyl group so we should have $(V^{\lambda})^*\simeq V^{\lambda}$ for any irreducible representation, i.e., taking the dual doesn't do anything.
So in particular, the linear part of the coordinate ring of $\mathrm{OG}(n,2n+1)$ is the $\mathfrak{g}$ representation $V^{\omega_1}$. Now, the linear part of this coordinate ring also seems like a perfectly good $G$ representation to me. And I would guess that it is the irreducible representation $V^{\omega_1}$. But that can't be right: $V^{\omega_1}$ should not be realizable as an $\mathrm{SO}(2n+1)$ representation, because of the fact that $\mathrm{SO}(2n+1)$ is not simply connected; to get this representation we are supposed to have to take the simply connected double cover $\widetilde{\mathrm{SO}}(2n+1)$, which is also called the spin group $\mathrm{Spin}(2n+1)$. (This representation $V^{\omega_1}$ is often called the spin representation.)
Question: where am I getting confused here? What is the coordinate ring of the orthogonal Grassmannian as a representation of the special orthogonal group?