I consider a closed area $ A $ and a piece of area $dA$ from it. Now let be $ \underline{n} $ the normal vector of $ dA $. Then we call $ d\underline{A}:=\underline{n}\cdot dA $ the area vector.
Now here are my problems:
1.) I don't see why $ \oint d\underline{A}=\underline{0} $.
2.) I struggle with this weird notation just writing the symbol $ d\underline{A} $ into to this integral. I'm used to use this notation like this for example $ \int f(x)\space dx $.
3.) Is there some (physical) interpretation for the integral $ \oint d\underline{A} $ ?
I assume in this context "closed" surface means the boundary is empty (and everything is nice enough for Stokes theorem to be applied). Now, one of the corollaries of Stokes' theorem is that for any smooth function $f$ and any "nice" surface $S$, \begin{align} \int_{S}\nabla f\,\times d\mathbf{A}&=-\int_{\partial S}f\,d\mathbf{l}. \end{align} Now, let $f_1(x,y,z)=x$, $f_2(x,y,z)=y$ and $f_3(x,y,z)=z$, so that $\nabla f_i=\mathbf{e}_i$ is the $i^{th}$ standard basis vector. Then, \begin{align} \mathbf{e}_i\times\left(\int_S\,d\mathbf{A}\right)&= \int_S\mathbf{e}_i\times d\mathbf{A}=\int_S\nabla f_i\times d\mathbf{A}= -\int_{\partial S}f_i\,d\mathbf{l}=0, \end{align} wherre the last equal sign is because the boundary $\partial S$ is empty. Now, it is easy to verify that if a vector is such that its cross product with each vector in a basis vanishes, then the vector must be to zero vector. Hence, $\int_Sd\mathbf{A}=0$.
The notation $d\mathbf{A}$ for example $\int_SF\,d\mathbf{A}$ means you're integrating the vector-valued function $F\,\mathbf{n}$ with respect to the area element on the surface $S$ (and if you want to get really technical, this is the induced Riemann-Lebesgue measure from the metric tensor induced on $S$ by pullingback the metric tensor in the ambient $\Bbb{R}^3$). Take a look at this answer of mine where I talk about various ways to interpret double integrals; granted they're mostly in the case where the function being integrated is real-valued, not vector-valued, but hopefully that is still helpful.
As we just showed, it is the zero vector, so I'm not sure what else you're looking for. It should seem like a plausible result though, atleast for highly symmetric surfaces like $S$ being the 6 sides of the cube (because the area vectors of opposite sides point in opposite directions, hence cancel out in pairs). Or for the sphere of any given radius $S_r=\{\xi\in\Bbb{R}^3\,:\,\lVert \xi\rVert=r\}$ (because in this case, at any point, there is a diametrically opposite point which will cause the normal vectors to "cancel out").
Note by the way that in proving $(1)$, if we make the additional assumption that $S$ is itself the boundary of some volume $\Omega$ (i.e $S=\partial \Omega$), then a simpler proof can be given by directly applying the divergence theorem: for any basis vector $\mathbf{e}_i$, we have \begin{align} \mathbf{e}_i\cdot \left(\int_Sd\mathbf{A}\right)&=\int_{\partial \Omega}\mathbf{e}_i\cdot d\mathbf{A}= \int_{\Omega}\text{div}(\mathbf{e}_i)\,dV=0. \end{align} Once again, if the inner product of a given vector with respect to every vector in a basis vanishes, then that vector must be the zero vector. This method of proving assumes more (that $S$ is itself the boundary of some region), but the proof is simpler, because it uses the regular form of the divergence theorem, rather than in (1) where we used a corollary of Stokes' theorem (which many of us may not even remember... I certainly had to look it up to refresh my memory on the signs).