Suppose I have the following, the gaussian surface is the drawing in the middle. So charge enclosed is zero, and then eletric field must be zero since the area of the gaussian surface is not zero. But, clearly the eletric field is not zero in the middle, because if you put a charge there it will move. Why I'm getting the concept wrongly? Edit: you can consider the middle figure as a sphere.
Gauss law says that the total flux going through a closed surface is equal to the charge inside the surface. You can think of flux as the number of field lines going in (or out) through the surface.
In your example there is no charge inside the sphere so the total flux through the surface of the sphere is zero. On one end (the left hand side) the field lines go in (negative flux) and on the other end (the right hand side) the field lines go out through the surface (positive flux). Gauss law in this case says that we have as many field lines going in as going out through the surface.
If the charge between the two surfaces is $0$ (zero) initially, that makes ${q}/{\epsilon} = 0$ and Gauss Law states that the Electric Flux is directly proportional to charge inside it, i.e.,
$\phi = \dfrac{q}{\epsilon} = \displaystyle\oint\vec{E}.\vec{dA}$ ,
Here $\vec{E}.\vec{dA}$ represents the dot product of the differential area and the electric field passing through it.
Even, if the flux is zero, $\displaystyle\oint\vec{E}.\vec{dA}$ becomes zero. Only if electric field and its area vector was always parallel i.e., always perpendicular to the surface, then the electric field can be claimed to be $0$ (zero).
