I was looking at the following paper by Tango:

In Lemma 2.4 , at the end, he says that if $Y \subset \mathbb{G}(k,n)$ is a subvariety of the Grassmannian of $k-$planes in $\mathbb{P}^n=\mathbb{P}(V)$ such that for every $y \in Y$ the corresponding $k-$plane $L_y \subset \mathbb{P}^n$ passes through a fixed point $p \in \mathbb{P}^n$ then the restriction of the universal bundle $$S=\{(x,L_x)|\:x \in \mathbb{G}(k,n)\} \subset V \otimes \mathcal{O}_{\mathbb{G}(k,n)}$$ to $Y$, i.e. $S_{|Y}$, has a trivial subbundle as a direct summand.

I understand that the fact that every $L_y$ passes through a fixed point $p$ gives me a section of $S_{|Y}$ simply by associating to every $y \in Y$ the point $p \in L_y$. This gives me the exact sequence $$0 \rightarrow \mathcal{O}_Y \rightarrow S_{|Y} \rightarrow S' \rightarrow 0$$ where $S'$ is a vector bundle with $rank(S')=rank(S_{|Y})-1$.

But I'm not able to see why this implies that $S_{|Y}=\mathcal{O}_Y \oplus S''$ for some $S''$.

Am I missing something? Thanks in advance for the help.