Let $M=\lbrace (x,y)\in S^2\times\mathbb R^3\vert \langle x,y\rangle=0\rbrace$. I proved it is a 4-dimensional submanifold of $S^2\times\mathbb R^3$ and that, for $\pi:(x,y)\in M\mapsto x\in S^2$, which is $\mathcal C^{\infty}(M,S^2)$, the set $\pi^{-1}(x)$ is a 2-dimensional $\mathbb R$-vector field for all $x\in S^2$. I want now to prove that every smooth map $\sigma:S^2\rightarrow M$ induce a vector field over $S^2$. I have the following definition of vector field: a map $X:S^2\longrightarrow TS^2=\displaystyle\coprod_{p\in S^2}T_pS^2$ such that:
$X_p\in TS^2$, and $ \forall f\in \mathcal C^{\infty}(S^2),$ $ Xf:p\in S^2\mapsto X_pf$ is smooth.