Pick an even dimensional real vector space $V$ of dimension $n$ and fix a symplectic form $\omega$ on $V$. Look at it as a map $\omega:V\otimes V\to\mathbb R$ by extending it from $\Lambda^2V$ to $V\otimes V$ as zero on the symmetric part. Fix also an inner product $\langle\mathord-,\mathord-\rangle$ in $V$.

Let $k$ be odd and such that $1\leq k\leq n$.

Let $W\subseteq V$ be an $k$-dimensional subspace, and let $W^\perp$ and $W^{\perp\omega}$ be the subspaces orthogonal to $W$ with respect to $\langle\mathord-,\mathord-\rangle$ and to $\omega$, respectively. We know that $\dim W^\perp=\dim W^{\perp\omega}=\dim V-\dim W$.

The restriction $\omega|_{W\otimes W^\perp}:W\otimes W^\perp\to\mathbb R$, which I will write for simplicity just $\omega_W$, is not zero. Indeed, if it *were* zero, we would have that $W^\perp$ is contained in $W^{\perp\omega}$, so in fact these two orthogonal subspaces would be equal, and in consequence we would have that $W\cap W^{\perp\omega}=W\cap W^\perp=0$. This would tell us that $W$ is in fact a symplectic subspace of $V$, which is absurd because it is odd dimensional.

Now $\omega_W$ is an element of $\hom(W\otimes W^\perp,\mathbb R)$, which identifies canonically with $\hom(W,\hom(W^\perp,\mathbb R))$. The inner product $\langle\mathord-,\mathord-\rangle$ restricts to an inner product on $W^\perp$ which allows us to identify canonically (because the inner product is fixed!) $\hom(W^\perp,\mathbb R)$ with $W^\perp$. After all these identifications, we have a non zero vector $\omega_W$ in $\hom(W,W^\perp)$.

Now, as explained in an answer to a MO question, $\hom(W,W^\perp)$ parametrizes a neighborhood of $W$ in $G(n,k)$, so it also can be identified with the tangent space to $G(n,k)$ at $W$.

We thus see that the rule $W\mapsto \omega_W$ gives a non-zero tangent vector field.