I need help with understanding the final reasoning in the proof of the Transversality Homotopy Theorem (Theorem 6.36) from John Lee's Introduction to Smooth Manifolds. Here, the goal is to show that the smooth map $F: N \times S \to M$ defined by $$F(p,s)=r(f(p)+e(p)s)$$ is a smooth submersion. So we need to show that for each $(p,s) \in N \times S$, $dF_{(p,s)}:T_p N \times T_s S \to T_{F(p,s)} M$ is surjective.
He does this by saying that for each $p \in N$, the restriction of $F$ to $\{p\} \times S$ is the composition of the local diffeomorphism $s \mapsto f(p)+e(p)s$ followed by the smooth submersion $r$, so $F$ is a smooth submersion.
I don't quite get how this proves surjectivity of the differential of $F$. To consider differentials, we need to look at $F$ in a neighborhood of $(p,s)$, so why does it suffice to restrict $F$ to $\{p\} \times S$?
What this seems to show is that for each fixed $p$, by the diffeomorphism $\{p\} \times S \approx S$, if we set $F_p$ to be the map $F$ restricted to $\{p\} \times S$, then $d(F_p)_s: T_sS \to T_{F(p,s)}M$ is surjective. But how do we guarantee that $d(F_p)_s = dF_{(p,s)}$?
A submersion has full rank. Suppose $$ F : R^k \times R \to R^p $$ is differentiable, and $F'(a, b)$ has rank $p$. Then $F$ is a submersion at $(a, b)$, right?
Now let $G(x) = F(x, b)$. Then $G$ is a map from $R^k \to R^p$.
Suppose I tell you that $G'(a)$ has rank $k$. What can you tell me about the rank of $F'(a, b)$? Well, it's at least the rank of $G'(a)$ because $$ F'(a, b) = \pmatrix{G'(a) & * \\ * & *} $$ so at least $k$ of its columns (the ones containing $G'(a)$ are linearly independent.
In Lee's case, we know that $G'(a)$ has rank $p$ (for every choice of $b$!). Hence $F'(a,b)$ has rank $p$ for every $(a,b)$.
So that's the proof for euclidean space, but for your manifold, everything is locally euclidean, so the chain rule finishes out the argument.