Let $\varphi: N \rightarrow M$ be $C^{\infty}$ and let $X$ be a $C^{\infty}$ vector field on $N$. Suppose that $d\varphi(X(p))=d\varphi(X(q))$ whenever $\varphi(p)=\varphi(q)$. Is there a smooth vector field $Y$ on $M$ which is $\varphi$-related to $X$?
This is problem 20 in Chapter 1 of F. Warner's Manifolds, and unfortunately, it is one of the few that has me stumped. I tried to consider simple cases such as when $M = \mathbb{R},$ $f$ the inclusion map, and $N$ some interval in $\mathbb{R}$ to get an embedding. But that didn't lead to much. Are we supposed to find a general counterexample if it is false? Or perhaps it is true — in any case, I would appreciate if someone could give me a step in the right direction.
Here's a counterexample, using the idea you described in your post.
Take $N = (0,\infty) $ and $M = \mathbb R$, and take $\varphi: N \to M$ to be the natural embedding $t \mapsto t$. Take $X$ to be the vector field $(1/t) \partial_t$ on $N$.
Because $\varphi$ is injective, the condition that $\varphi(t_1) = \varphi(t_2)$ implies $d\varphi(X(t_1)) = d\varphi(X(t_2))$ is satisfied vacuously.
Now if $Y$ is a vector field on $M$ that is $\varphi$-related to $X$, then $Y(t) = (1/t)\partial_t$ for all $t > 0$. But then, $Y(t)$ cannot possibly be continuous at $t = 0$, and we have a contradiction.