Since this question has remained open for a while, I thought I should think more carefully about what's involved. @Ted Shifrin's suggestion is the right way to start, but after thinking about it for a while, I decided that the problem really needs a better hint than the one I put in the book. So here's a (hopefully) better suggestion to get you started. (I've also added this to my list of corrections on my ISM web page.)
First, to get a good result,
you'll have
to add the assumption that $\ker dF_p\not\subseteq T_p\partial M$.
After choosing smooth coordinates, you can assume $M \subseteq \mathbb H^m$
and $N\subseteq\mathbb R^n$, and extend $F$ to a smooth function $\widetilde F$ on an
open subset of $\mathbb R^m$.
Now, as Ted suggested,
assuming that $F$ has constant rank $r$, show that there is a coordinate projection $\pi\colon\mathbb R^n\to\mathbb R^r$ such
that $\pi\circ \widetilde F$ is a submersion, and apply the rank theorem
to $\pi\circ \widetilde F$ to find new coordinates in which $\widetilde F$ has a coordinate representation
of the form $(x,y) \mapsto (x,R(x,y))$.
Then use the rank condition to show that $R|_M$ is independent of
$y$.
Consider a smooth function $f\colon M\to N$. If $X$ is an arbitrary manifold, then the map $F\colon X\times M\to X\times N$ given by $F(x,p)=(x,f(p))$ is a fiber bundle homomorphism. But if $f$ is smooth, bijective, and not a diffeomorphism, then $F$ is smooth, bijective homomorphism, but it isn't an isomorphism.
In order to give an specific example: take the smooth manifold $\mathbb{R}$ and think of $\mathbb{R}\times \mathbb{R}$ as a trivial bundle. A possible counterexample can be $F(x,y)=(x,y^{3})$.
The reason the argument works for a vector bundle is that the bijective condition implies that, in fibers, $F$ is a linear isomorphism (hence a diffeomorphism). When you remove the vector structure, you only know that it is a smooth bijective map on the level of fibers.
FOR THE CASE WHEN OUR BUNDLES ARE VECTOR BUNDLES
Suppose $F\colon E\to E'$ is a vector bundle isomorphism over $f\colon M\to M'$. This means that there is a vector bundle homomorphism $G\colon E'\to E$ over some smooth function $g\colon M\to M'$ such that $G=F^{-1}$ as maps. In particular, $g=f^{-1}$, so $f$ is a diffeomorphism. On the other hand, $F_{p}\colon E_{p}\to E'_{f(p)}$ is a linear isomorphism with inverse $G_{f(p)}\colon E'_{f(p)}\to E_{p}$.
Now suppose $f$ is a diffeomorphism and $F_{p}$ is an isomorphism for every $p$ (in the case that $F$ is bijective, the second part is true). Then $F$ is bijective, and by using local trivializations you can see that $F_{*}$ has maximal rank at every point of $E$, so that $F^{-1}$ is a diffeomorphism (and a homomorphism).
EDIT: For completeness sake, I'm adding a proof of the fact that $F$ is a local diffeomorphism.
Let $p\in M$, $U\subseteq M$ and $V\subseteq M'$ trivial coordinate open subsets for $E$ and $E'$, with coordinate systems $\varphi\colon U\to \varphi(U)\subseteq \mathbb{R}^{n}$, $\psi\colon V\to \psi(V)\subseteq \mathbb{R}^{n}$ such that $f(U)\subseteq V$, let $\pi$, $\pi'$ be the projections, and $\alpha\colon E_{U}\to U\times \mathbb{R}^{k}$, $\beta\colon E'_{V}\to V\times \mathbb{R}^{k}$ trivializations for both bundles (with second components $\alpha_{2}$ and $\beta_{2}$). Let $\tilde{f}\colon \varphi(U)\to \psi(V)$ be the coordinate representation of $f$ (that is, $\tilde{f}=\psi\circ f \circ \varphi^{-1}$). Then $E_{U}$ and $E'_{V}$ are coordinate open subsets with the maps $\Phi\colon x\in E_{U}\mapsto (\varphi(\pi(x)),\alpha_{2}(x))$ and $\Psi\colon z\in E'_{V}\mapsto (\psi(\pi'(z)),\beta_{2}(z))$. A coordinate representation of $F$ with respect to $E_{U}$ and $E'_{V}$ is therefore given by $\tilde{F}=\Psi\circ F \circ \Phi^{-1}\colon \varphi(U)\times \mathbb{R}^{k}\to \psi(V)\times \mathbb{R}^{k}$.
We now compute $\tilde{F}$: let $(x,v)=(x^{1},...,x^{n},v^{1},...,v^{k})\in \varphi(U)\times \mathbb{R}^{k}$. We get:
$$
\tilde{F}(x,v)=\Psi \circ F \circ \Phi^{-1}(x,v)=(\psi(\pi'(F(\Phi^{-1}(x,v)))),\beta_{2}(F(\Phi^{-1}(x,v))))=(\psi(f(\pi(\Phi^{-1}(x,v)))),\beta_{2}(F(\Phi^{-1}(x,v))))=(\tilde{f}(x),\beta_{2}(F_{\varphi^{-1}(x)}(\Phi^{-1}(x,v))))=(\tilde{f}(x),A(x)v)
$$
where $A(x)=\beta_{2}\circ F_{\varphi^{-1}(x)}\circ \Phi^{-1}(x,\cdot)$ is a linear isomorphism by hypothesis. The Jacobian matrix is therefore
$$
D\tilde{F}(x,v)=\begin{pmatrix}
D\tilde{f}(x) & 0 \\
* & A(x)
\end{pmatrix}
$$
which is nonsingular since both diagonal blocks are nonsingular.
Hope this helps!
Best Answer
Perhaps this will help build a bit of intuition. The theorem your citing is about regular points on manifolds and their preimages. The key here is that the theorem says nothing about how these preimages vary as the point moves smoothly on the manifold. With the kernel of a bundle map, ensuring that it is a submanifold involves showing that the sub spaces smoothly vary as we vary the point p. One can see that the only way we can get smoothness is by looking at the smooth structure of the base space of the target bundle. This is exactly what the theorem your citing is missing...smoothness in p