A vector bundle section with surjective vertical differential at every zero has a submanifold zero set

Let $p\in Z(s)$ and choose a local frame near $p$. Then $s(x)={\sum }_i{f}_i(x)e_i(x)$ for a smooth map $f=(f_1,\dots ,f_r):U\to {\mathbb{R}}^r$. Because $s(p)=0_p$, one has $f(p)=0$. Under a change of frame by a matrix $A(x)$, the new component map is $A(x)f(x)$, whose derivative at $p$ is $A(p)d{f}_p$ because the term $(dA{)}_{p}f(p)$ vanishes. Thus surjectivity of the vertical differential is exactly surjectivity of $d{f}_p$, independent of the chosen frame.

Near $p$, the zero set of $s$ is therefore the zero set of the component map $f$, and $0\in {\mathbb{R}}^r$ is a regular value because $d{f}_p$ is surjective. By [L2], $f^{-1}(0)$ is an embedded submanifold of codimension $r$. Doing this at every zero proves that $Z(s)$ is an embedded submanifold of codimension $r$.