I am asking about a slightly different version of this question, where we are given an embedded submanifold $M \subset M'$ and are asked to extend any smooth function on $M$ to one on a neighborhood of $M$ in $M'$. Assuming our function to be real-valued this is clear, a solution is given in the link. However, for an arbitrary codomain I am not quite sure if it can be done using very elementary techniques. I am able to prove it assuming the embedding theorem for general manifolds (Lee, if you are reading this: are you assuming the reader knows that any smooth manifold $M$ can be embedded in finite-dimensional Euclidean space?)
Under the assumption in the parenthesis the result is provable using the same technique. I have been unable to patch things together using charts as the differential structure on $N$ might be weird. Another technique mentioned when discussing with other people is that of tubular neighborhoods, however that theorem is given as an exercise in Chapter 8 in Lee and is unlikely to be assumed as a prerequisite for the introductory chapters.
Am I making my life too difficult here? Can I patch things together using charts?
OK, I see that I was unclear. In Exercise 2.3, I did not mean to ask about arbitrary codomains. When I said "smooth function," I was using the typical convention in differential geometry of reserving "function" to mean a real-valued or vector-valued function, not a function into another manifold. But I should have said that. I've added a correction to my errata. (Note also that there was already a correction that changed "extended to a smooth function on $\widetilde M$" to "extended to a smooth function on a neighborhood of $M$ in $\widetilde M$." That's now incorporated into the new correction.)
The question of extending smooth maps into arbitrary codomains does require more machinery. You can probably prove the existence of an extension on some neighborhood of $M$ by using some combination of tubular neighborhoods, the Whitney embedding theorem, and the Whitney approximation theorem. There's also a global result: If $M$ is closed in $\widetilde M$ and $f\colon M\to N$ is smooth, then there exists a smooth extension of $f$ to all of $\widetilde M$ if and only if there exists a continuous extension. (This is Corollary 6.27 in my Introduction to Smooth Manifolds, 2nd ed.).