Why is the image of a smooth embedding $f: N \to M$ an embedded submanifold?

The reverse inclusion doesn't necessarily follow from the canonical form theorem you've stated, since $V$ can just happen to be too large (or, if you like, $U$ can be too small): given $U$ and $V$ as you have one can always replace $U$ by a much smaller neighborhood $U'$ of $0$, and still guarantee that $U'$ and $V$ have charts satisfying your theorem -- and also be sure that the inclusion $\supset$ can't hold.

The solution, of course, is just to prove a slightly better canonical form theorem, by reducing the possible size of $V$:

Theorem. For any $p \in N$, there exist charts $(U, \phi = x^1, \dots, x^n)$ and $(V, \psi = y^1, \dots, y^m)$ satisfying $$ \psi f \phi^{-1} : (r^1, \dots, r^n) \mapsto (r^1, \dots, r^n, 0, \dots, 0) $$ and such that $$ f(U) = \lbrace q \in V : y^{n+1}(q) = \cdots = y^m(q) = 0 \rbrace. $$

Proof. The proof is simple once you have your canonical form theorem; if $(U, \phi)$ and $(V', \psi)$ are the charts guaranteed by your theorem, then you can take $$ V = \psi^{-1} \Big( \psi(V') \cap (\phi(U) \times \mathbb{R}^{m-n}) \Big). $$ Then $(U, \phi)$ and $(V, \psi|_V)$ satisfy the conditions of this new theorem.