How does $v\Phi^1=\cdots=v\Phi^k=0$ imply $v\in\ker d\Phi_p$?

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I'm confused about an immediate corollary in John Lee's Smooth Manifolds. Proposition 5.38 says

Suppose $M$ is a smooth manifold and $S\subset M$ is an embedded submanifold. If $\Phi\colon U\to N$ is any local defining map for $S$, then $T_pS=\ker d\Phi_p\colon T_pM\to T_{\Phi(p)}N$ for each $p\in S\cap U$.

It says an immediate corollary is that if $S\subseteq M$ is a level set of a smooth submersion $\Phi=(\Phi^1,\dots,\Phi^k)\colon M\to\mathbb{R}^k$, then $v\in T_pM$ is tangent to $S$ iff $v\Phi^1=\cdots=v\Phi^k=0$.

Why does $v\Phi^1=\cdots=v\Phi^k=0$ imply $v\in T_pS$? By the proposition, how does $v\Phi^1=\cdots=v\Phi^k=0$ imply $v\in\ker d\Phi_p$? I don't see how to relate this to the individual components though to use $v\Phi^1=\cdots=v\Phi^k=0$.

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Here's one way to prove it. If you write $d\Phi_p(v) = a^i \partial/\partial x^i$ (using the summation convention), you can compute each coefficient $a^k$ by letting $d\Phi_p(v)$ act on the $k$-th coordinate function $x^k\colon \mathbb R^n\to \mathbb R$: $$ a^k = d\Phi_p(v)(x^k) = v (x^k \circ \Phi) = v(\Phi^k). $$

Part of the reason you had a hard time following @Shuchang's answer is because he's probably thinking of $d\Phi^k_p$ as a covector, but in Chapter 5 I haven't yet introduced covectors. This will be easier to see after you've read the introduction to covectors in Chapter 11.

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It's redundant to consider $f$. A straightforward verification that since $v\in\ker d\Phi_p$, then $0=d\Phi_p(v)=v\Phi_p$ and hence $v\Phi^k=0$ for each $k$.