Obviously this doesn't hold if we replace "smooth" with something like "analytic" or "regular," which are the contexts I'm more familiar with. And obviously we can't extend a smooth function defined on an arbitrary open set to a global function. But so far as I can see it seems like we might well be able to extend an arbitrary germ to a global function.
I imagine this is a pretty dumb question from the perspective of someone who knows even a little bit of differential geometry, but I wasn't able to find an answer quickly by Googling.
Let $f$ be smooth around a point of manifold $M$. Via a map, we have a smooth function on a small open ball in $\mathbb R^n$. Using a standard construct of a smooth function that is $0$ outside that open ball and $1$ in a smaller ball, we can extend the smooth germ to all of $\mathbb R^n$ and hence to all of $M$ (taking the value $0$ outside tha map we started with).