I am supposed to calculate the electric field $E$ created by a electrical charge $Q>0$ distributed on the surface of a plane. For this I should use
(i) Gauss' theorem $$\int_M \operatorname{div}(E)dV = \int_{\partial M} \langle X , N\rangle dS,$$ where $dV$ denotes the volume form on some compact manifold $M$ and $dS$ the corresponding pulled-back volume form on $\partial S$ together with the surface normal $N$ on $\partial S$.
(ii) Gauss' law of electromagnetism $$ \operatorname{div}(E) = \rho $$ (I am omitting the physical constants.)
I actually don't see how this formalism applies in this case: Since the charge should be distributed on the surface (and only on the surface) of the plane how should I write down the charge density $\rho$? Since the plane is of measure zero in the 3-dimensional case, there can't be a $\rho$ on $M$ such that $$ \int_F \rho \ dV = Q$$ for all $F \supseteq S$, where $S$ is the plane in question and $F$ a suitable subset of $M$ over which the differential form $\rho dV$ can be integrated over.
Of course it is possible to write down a differential form on $S$ by $$ dQ := \frac{Q}{\int_S dS}dS, $$ where $dS$ is the volume form on $S$. Then $\frac{Q}{\int_S dS}$ is the charge density on $S$, but obviously the electric field distributed on the plane not only lives in $S$ but only on $M$. But Gauss law of electromagnetism doesn't say anything about surface charges. How can I solve this problem mathematically in the language of vector calculus / differential forms? I know that physisicists would probably employ some $\delta$-trickery, but to make this mathematically precise one would have to use some measure theory and I don't know how to apply Gauss' theorem when we're not integrating differential forms but some measures like $\lambda^2 \otimes \delta$.
Let me address the original problem of calculating the electric field of a charged plane. It is true that Gauss' Theorem may be written as you posted, but I think what you are looking for is Gauss' law of electrodynamics: $$\int_{S}\vec{E}\cdot \hat{n}dS = \frac{q}{\epsilon_{0}}$$ where I'm assuming your charged plane is in the vacuum. Here, $S$ is a Gaussian surface and $q$ is the total charge enclosed by $S$. Now, suppose a part of your (infinite) plane is enclosed by a parallelepiped. By symmetry, you expect to have electric field only through the upper and lower face of this parallelepiped. Moreover, the same symmetry argument can be used to conclude that the electric field must point upward and downward on the upper and lower face, respectivelly, and it could only be $z$ dependent. Thus, because $\hat{n}= \hat{z}$ on the upper face of the parallelepiped and $\hat{n}=-\hat{z}$ on the lower face, we have $\vec{E}\cdot \hat{n} = |\vec{E}|$ on both faces. If these faces have area $A$, the area of the plane enclosed by this parallelepiped is also $A$.If the total charge on this area $A$ is $q$, the surface density is $\sigma = q/A$. Finally, using Gauss' law: $$\int_{S} \vec{E}\cdot\hat{n}dS = 2EA = \frac{\sigma A}{\epsilon_{0}} \Rightarrow |\vec{E}| = \frac{\sigma}{2\epsilon_{0}}$$
Note: The assumption that the plane is infinite is just to avoid divergences and lack of symmetry. It does not play much role in the actual calculations.