Your answer has the word conductor in it, your question does not have the word conductor in it. If it is a conductor with some finite thickness you can place the pillbox to enclose just one surface of the conductor.
Yes, I agree, but this would still change the definition of surface charge from the one I quoted in the common Gauss's law result I quoted above, which is what I was getting at.
The definition of surface charge is a charge per area. If you want to make a surface so thin that you cannot slice it with a Gaussian surface then there are fields on both sides and hence charge on both sides. If it has thickness then you can get the surface charge on just one side by using Gauss's Law to enclose just one side. Something you did not do. I would think you understood except some books (usually more mathematical ones) do start with fields and then define charge from derivatives of fields. They do that to avoid stating exactly which space the charge lives in. I.e. if the fields live in H curl or in H div or in the intersection then you don't have to define what space the charge distribution lives in. In case your books are like that I want to be clear that surface charge already has a definition.
All I'm saying is that in the common Gauss's law derivation, "surface charge density" has a different meaning than in other cases that you analyze, and that because of this the Gauss's law relationship can be misleading.
Whether you want to say this is because the Gauss's law surface is the "only" surface in the Gaussian surface or because the operational definition in the Gauss's law is the surface charge on "both" surfaces seems to be a matter of semantics.
Let's be clear. 1) Gauss's Law is correct. 2) Gauss's Law only tells you the charge enclosed inside a volume of your choosing 3) It is up to you to place the Gaussian surfaces around an actual physical surface to learn anything about that physical surface. When you have two physical surfaces then each produces an electric field so the field is twice as strong because it is due to both. You can get this by superposition. This really happens in a conductor. The field is zero on the inside of the conductor because of the surface charge on both sides of the conductor contribute 4) A Gaussian surface isn't inside a Gaussian surface. 5) Gauss's Law isn't misleading it relates charge enclosed to electric flux 6) When you see an electric field outside a conductor of finite thickness there really is half as much charge on that one surface as you might expect because you really can fit a Gaussian surface inside the body of the conductor. And because the field is half of it due to other charges on the other parts of the surface. They have to be strong enough to produce an equal and opposite field just on the inside so just on the outside they produce an equal and equal field. This is a real effect, not a semantics issue.
Edit If it helps, you can think of the first rule $\sigma = \epsilon_0 \partial V/\partial n$ as a special case. The special case when the electric field right at the surface is only due to the surface charge right there. For a conductor the surface charge right there is only responsible for exactly half the field right there (on the one side of it).
Gauss's Law is always correct. And it is not misleading, you need to use it to learn what it is saying. The equation $\sigma = \epsilon_0 \partial V/\partial n$ is misleading because it only applies sometimes and gives a false sense of generality. It would make you think there is charge in between the plates of a parallel plate capacitor just because there is an electric field there. No one thinks that there is surface charge just because there is an electric field.
You can have some field on two sides of a surface, and then jump indicates some surface charge. This would be the proper version of the $\sigma = \epsilon_0 \partial V/\partial n$ rule.
So the correct rule is $\sigma = \epsilon_0\left( \hat n_1\cdot \vec E_1+\hat n_2\cdot \vec E_2\right)$ where you evaluate on both sides of the surface and the unit normal is always outwards. The jump in the normal component is proportional to the surface charge. Just like Gauss says. And this resolves the senseless sign issue in your version.