Let's start with $x \in Y$ a $B$-fixed vector. Then it's a high weight line in $M$, so the linear span $N \leq M$ of $Y$ is an irreducible subrepresentation of $M$.

Do you mind if I assume we're in characteristic $0$, as Brion does? Then we can split $M = N \oplus N'$, and $M^* = N^* \oplus N'^*$.`
(Which is to say, you can reduce to the case $M$ irreducible, if that helps understand the situation.) So it's sufficient (and in the general case, necessary!) to look for your $\eta$ inside $N^*$.

Your first question is then why the $\eta \in (N^*)^B$, unique up to scale, has $\eta(m) \neq 0$. If it were zero,
$$ 0 = \eta(m) = (b\cdot \eta)(m) = \eta(b^{-1}\cdot m) \qquad \forall b\in B $$
from which we learn that $\eta$ annihilates a spanning set of $N$, so all of $N$. But $\eta \in N^*$, ` so it's zero, contradiction.

I'm not used to describing parabolic subgroups as "opposed" but I assume it means that $G_{[\eta]} G_y$ is dense in $G$. By definition of $\eta$, $G_{[\eta]} \geq B$, so it's enough to show $B G_y$ is dense in $G$, or that $B G_y / G_y$ is dense in $G/G_y$. This was exactly the condition used to pick $y$, that $B y$ was dense in $Y \cong G/G_y$.