Let $M, N$ be smooth manifolds s.t. $N$ has no boundary. Let $A\subset M$ be a closed subset. $F:A\rightarrow N$ is called smooth if there is an extension $F':U\rightarrow N$ for some open $U\subset M$.
But can we always extend a smooth $F$ to the whole $M$ ?
I know that there is a smooth extension iff there is a continuous extension. Also, if $N=\mathbb{R}^k$ then the answer is positive.
Smoothness doesn't actually come into it. As you noted, continuous extension is the main obstruction. There are at least two things to worry about.
Connectedness (as mentioned by Moishe Kohan in the comments).
Suppose that $M$ is connected, $A = A_1 \sqcup A_2$ has multiple components, and $N = N_1 \sqcup N_2$ also has multiple components, and $F$ is such that $F(A_1) \subset N_1$ and $F(A_2) \subset N_2$, then there cannot be any continuous extension of $F$ that maps $M \to N$.
Topology
Continuous maps induce (for example) homomorphisms of the fundamental groups. This can cause some trouble. For example, let $M = \mathbb{S}^2$ and $N = \mathbb{S}^1$. Let $A$ be the equator and $F$ the obvious mapping. If there were a continuous extension of $F$ to the whole manifold, then since $M$ has vanishing $\pi_1$ the image of every closed curve under $F$ has to be contractible in $N$. However, this is manifestly not the case given our prescription that the equator is mapped into $N$ as a non-contractible loop.