@Thor: First, I assume $\pi(x,e_1,\dots,e_n) = (x,e_1)$. So, let's see. $\pi\circ s(x) = e_1$, so you're asking for $(de_1)_p(e_j) = \sum_{k=2}^n \omega_1^k(e_j)e_k$, all evaluated at $p$. Note that $e_2,\dots,e_n$ span the tangent space of $SM$ at $(x,e_1)$.
@MikeMiller I resemble that remark. My best friend in San Diego and I always say "irregardless" to one another ... as sarcastic humor. It's funny when other friends correct us, not realizing apparently that we of course know better.