I basically have this example in mind. Consider the unit circle $S^1$ as a submanifold of $\mathbb{R}^2$ (symmetric about the origin) and $[-1,1]$ as a submanifold of $\mathbb{R}$. Define the projection map $\pi: S^1 \to [-1,1], \pi(x,y) = x$ where $(x,y) \in S^1$. This map is smooth in the induced smooth structures and has constant rank 1 on $S^1$ (rank of $d\pi_{(x,y)}$ is 1). So by the inverse function theorem, every $x \in [-1,1]$ is a regular value and so there are open neighborhoods about $x$ and any point in its fiber that are diffeomorphic. But if $x = -1$ then there appears to be a problem because the open neighborhoods of $x$ are of the form $[-1,a), a\leq 1$ and $\pi$ is not injective on the preimage of any such open set.
So either I've made a mistake above or there can be issues at ramification points of the smooth map.
The projection map $\pi: \Bbb R^2\to\Bbb R$ defined by $(x,y)\mapsto x$ has indeed rank $1$, but this doesn't imply that the projection map $\pi: S^1\to \Bbb R$ (considered as a map between these manifolds) has rank $1$ at every point. Indeed, in $(\pm 1,0)$, it has rank zero.
If you parametrize a neighborhood of $(1,0)$ in $S^1$ by a parameter $\varphi\in (-\epsilon, \epsilon)$ and a map $\varphi\mapsto (\cos \varphi, \sin \varphi)$, then $\pi$ is represented (in this coordinate frame) by $\varphi\mapsto \cos \varphi$, which is a function with zero derivative for $\varphi=0$.