For Gauss' law to be applied straightforwardly, your point of interest must lie on an equipotential surface with some symmetry. For the case of the point inside the sphere, it is easy to visualize a Gaussian sphere with radius $r<R$ with said point on its boundary. The charge contained inside the Gaussian sphere would be zero -- the flux through the Gaussian sphere would be zero. You can then argue that, since the field lines should be normal to this Gaussian sphere, the flux $\phi_E = 4 \pi r^2 \textbf{E} \cdot \hat{r} = 0$ implies $\textbf{E} = 0$ inside.
It's quite different outside the charged sphere, however. If you try to hypothesize a similar Gaussian surface, you'll fail, because the electric field due to the charged sphere varies with position, and no Gaussian sphere would be normal to the field lines, unless you envelop the charged sphere completely (which would amount to computing the field at $r>R$.) The summary is that you cannot equate the flux to zero and infer the field is zero as well, like in the previous case. The net flux would be zero, sure, but that would involve a complicated mess of integrals and dot products with the radial vector and the normal vectors of our Gaussian surface.
Perhaps it is more illuminating to look at what is actually happening to the spherical shell, which we'll assume is perfectly conducting. In the absence of an external electric field, the free charges in the shell will align themselves such that the net electric field in the shell is zero; otherwise, they will simply move around until that occurs. On the other hand, if the shell is placed in an external field $\textbf{E}_\text{ext},$ this will prompt the free charges in the shell to reorient themselves such that the induced field $\textbf{E}_\text{ind}$ cancels out the external field perfectly everywhere inside the shell: $$\textbf{E}_\text{ext} + \textbf{E}_\text{ind} = 0$$
I hope this clears your doubt.