This has basically already been answered, but because all your boundaries are smooth (or, more importantly, $C^1$), and your domains bounded, then if you want an explicit smooth continuation, you could do the following: for a sufficiently small $\epsilon > 0$, you can construct a subset $V \subset \Omega'$ such that for any $x \in \partial V$, dist($x,\partial \Omega'$) = $\epsilon$$\operatorname{dist}(x,\partial \Omega') = \epsilon$. Then, "radially" along these lines of length $\epsilon$ connecting the boundary of $V$ to the boundary of $\Omega'$, you can have your continuation of $f$ smoothly vanish to 0 using a scaling of the form $e^{-1/x}$. Then set $f = 0$ on $V$.
Therefore this continuation of $f$ will "mostly" vanish in $\Omega'$. You're basically just constructing a smooth mollification of $f \chi_{\Omega \setminus \Omega'}$.
The problem with this construction is that, from an application point of view, it's not very helpful. By making $f$ vanish on the interior set, you've basically lost all the information that was encoded in $f$ in the outer set.
