Let us work over $\mathbb C$, using the Grothendieck projectivization $\mathbb P():=Proj(Sym())$.
Consider a $7$-dimensional vector space $V$ endowed with a symmetric non-degenerate bilinear form $q:V \times V^\vee \to \mathbb C$. We define $$ OG(2,7):=\{\mathbb P W \subset \mathbb P V: \dim W=2 \text{ and }W \text{is }q-\text{isotropic}\}. $$ Consider the tautological bundle sequence $$ 0 \to \mathcal Q^\vee \to V^\vee \otimes \mathcal O \to \mathcal S \to 0 $$ where $\mathcal Q^\vee$ is the tautological subbundle over $OG(2,7)$. Thanks to the quadratic form, we can define the orthogonal bundle $$ \mathcal Q^\vee_\perp \to OG(2,7) $$ that is a rank $5$ vector bundle over $OG(2,7)$ which over a point $[\mathbb P W]$ it puts the vector space $$ W^\vee_\perp=\{v \in V: q(v,w)=0 \text{ for all }w \in W^\vee\}. $$ Using the identification given by the bilinear form, $V \simeq V^\vee$ and we can decompose $$ V^\vee=U \oplus Z \oplus Z' $$ where $\dim U=1$ and $\dim Z=\dim Z'=3$. In particular, $W^\vee \subset Z$ and $W^\vee_\perp$ is the orthogonal complement. One can show that $U \subset W^\vee_\perp$, essentially by looking at what happens in the even case.
This last fact allow us to write $\mathcal Q^\vee_\perp$ as the central term of a short exact sequence over $OG(2,7)$ $$ 0 \to \mathcal O \to \mathcal Q^\vee_\perp \to \mathcal E \to 0 $$ that correspond fiberwise to $$ 0 \to U \to W^\vee_\perp \to W^\vee_\perp/U \to 0. $$
Question: The space of global sections $\Gamma(OG(2,7),\mathcal E)$ corresponds to a representation of $SO(7)$, which one? I hope that it is the tensor product between $V$ and the $8$-dimensional spin representation.
I think that one should use the Borel-Weyl_bott theorem, by I frankly never see an explicit computation, so a step-by-step resolution would be wonderful.
