I don't work with smooth manifolds, but in algebraic geometry this could never happen, at least not for a coherent sheaf (which should be the case if your vector spaces are finite dimensional).
For simplicity let's assume that $K=\{x\}$, that is, we're in case (2).

Grothendieck's theorem on local cohomology says that those sheaves/groups vanish if $t<i<d$, but do not vanish when $i=t$ or $i=d$ where $t$ is the depth and $d$ is the dimension of $F_x$ as a module over the local ring $\mathscr O_{X,x}$.

I realize this might not cover your situation but I'd say that it suggests that you should not expect a blanket vanishing theorem like that.

I also offer the following experiment that might show that the same phenomenon is happening in your case:

Suppose $X$ and $F$ are such that $F$ has no higher cohomology on $X$, i.e.,
$$
H^i(X,F)=0, \quad\text{ for $i>0$}. \tag{$\star$}
$$
I suppose the euclidean space with a trivial bundle is such. Now, I would expect that it should be easy to find a compact subset $K\subset X$ such that the same does not hold for $X\setminus K$ and $F\left|_{X\setminus K}\right.$. Again, in algebraic geometry this is easy: Choose $X$ to be an arbitrary affine variety (say a euclidean space) of dimension at least $2$, $F$ a trivial bundle and $K=\{x\}$ where $x\in X$ is a smooth point.

So, if you have that setup, then consider the long exact sequence of local cohomology. By the assumptions, your condition cannot hold. (Because it would imply the same vanishing as in $(\star)$ on $X\setminus K$.)