For simplicity I'm considering only the sphere case.
In the Gauss' Law formulation we have some field $E$ introduced by charges $Q$ inside some sphere, then we compute flux and integrate, and we get result $Q/e_0$. Right.
But this formulation doesn't take into account any possible outer charges, because $E$ used in this law comes only from $Q$. I suppose that any outer field effects will cancel out, because flux coming from it must go through surfaces with opposite normals. But we didn't make any proof of that, as far as I have seen in Gauss' Law proofs.
So, when we measure electric field inside the sphere with uniformly distributed $Q$, we make some assumptions on uniformity of $E$ inside this sphere and we consider hollow Gauss' sphere with radius smaller than radius of the outer sphere. And then we use $E$ to compute total flux going through inner sphere and conclude that it must be zero. Right.
But why can we use this $E$? We don't know that there aren't any effects from outer charges fields. So why can we use Gauss' Law?
Best Answer
The electric field used in Gauss's law is not just the electric field from the interior charge distribution. It may contain contributions from exterior sources. Usually, we do assume that there are no exterior sources, which is what allows us to impose symmetry considerations.
Why are we able to do this? Because Maxwell's equations are linear, and as such, the sum of two solutions to Maxwell's equations is itself a solution. Hence, you can consider the exterior and interior charge densities separately. You can calculate the fields separately and then add them up at the end.
Here's a mathematical way of looking at it. You have Gauss's law in integral form:
$$\int_V \frac{\rho}{\epsilon_0} \, dV = \oint_S E \cdot dS$$
Linearity of Maxwell's equations means that you can split up $\rho = \rho_\text{int} + \rho_{\text{ext}}$ and the same for $E$. The two parts must separately obey this equation. So that means we have
$$\begin{align*} \int_V \frac{\rho_\text{int}}{\epsilon_0} \, dV &= \oint_S E_\text{int} \cdot dS \\ \int_V \frac{\rho_\text{ext}}{\epsilon_0} \, dV &= \oint_S E_\text{ext} \cdot dS \end{align*}$$
But $\rho_{\text{ext}} = 0$ everywhere in $V$, so the second RHS integral must be zero.
When we do Gauss's law problems, we implicitly look at the $E_\text{int}$, the field generated by charges internal to the volume. Any external field and charges can be separated out in this way, so that the symmetry of the field generated by internal charges is still manifest, allowing us to compute the field magnitude thanks to that symmetry. It's true that, strictly speaking, we can't draw any conclusions about the total $E$-field without some information about exterior charges or fields, but by linearity, we know that our field from the internal charges will never be affected by such information.