What exactly does the line integral of an electric field over a closed loop find? For a conservative electric field, we can always say that $$\oint \vec E\cdot\vec {dl} = 0$$
Take this scenario for example

Here $E_1$ and $E_2$ are two different electric fields. If I find
$\oint \vec E\cdot\vec {dl}$ from $A$ to $A$(loop), then I know it is $0$, because the potential difference between $A$ and $A$ will always be zero (since they are the same point).
Now, here only vertical lines (PQ and RS) would be nonzero.
So,
$$\oint \vec E\cdot\vec {dl} = 0$$
$$\int_P^Q \vec E_1\cdot\vec {dl} + \int_S^R \vec E_2\cdot\vec {dl} = 0$$
$$E_1\cdot l - E_2\cdot l = 0$$
$$E_1 = E_2$$
But then, it contradicts my assumption of $E_1$, $E_2$ being different.
It gives rise to two possibilities:

*

*The potential difference between $A$ and $A$ is not zero.

*You can never create two different electric fields, because if you do, I could always draw a closed loop encapsulating both the fields and prove that the fields are same.

So what is going wrong here?
 A: You have created a vector field that cannot be an electrostatic field. This is because $\oint \mathbf E \cdot \text d\mathbf l \neq 0$ for the field.
If you want to be more careful with your setup, finite lines of charge have fringing fields that extend past the lines of charge. If you were to take this into account then you would get a zero line integral.
A: Consider a smooth vector field $\mathbf{E}:\Bbb{R}^3\to\Bbb{R}^3$ of the form $\mathbf{E}(x,y,z)= f(x,z)\,\mathbf{e}_y= (0,f(x,z),0)$. Now, by integrating over a rectangular loop like yours (lying inside a plane of constant $z$) we find that
\begin{align}
\int_{\text{loop}}\mathbf{E}\cdot d\mathbf{l} &= (E_1-E_2)L
\end{align}
So, if $E_1\neq E_2$ then the RHS is non-zero which proves for you that such a vector field $\mathbf{E}$ is NOT conservative.
To answer your two questions explicitly:

*

*If you take an electrostatic field (which is conservative pretty much by definition of "electrostatic") and integrate over a closed loop, then the result is always zero. Hence it is trivially true that the potential difference (which is well-defined due to the field being conservative) between point $A$ and point $A$ is $0$.


*"You can never create two different electric fields..." You should be very careful with your wording. The correct statement is "a smooth vector field of the form $\mathbf{E}(x,y,z)=f(x,z)\,\,\mathbf{e}_y = (0,f(x,z),0)$ (where $f$ is a non-constant function of $x$) is not conservative, and hence does not arise as a result of an electrostatic field."

The reason you got confused is because you let the drawing deceive you. You seem to draw two parallel plate capacitors, which is the typical example of a charge configuration which produces a constant field in the direction normal to the plates. However, you should keep in mind that this is only true when the plates are infinite in size (so we certainly can't have two sets of these "side by side").
In the case you've drawn, there are two finite-in-size parallel plate capacitors. In this case, the electric field DOES NOT have the special form $\mathbf{E}(x,y,z)= (0,f(x,z),0)$, which is what is erroneously suggested by the drawing. The correct field looks very complicated. Below is a picture I found which roughly shows what the field lines look like for a single parallel-plate capacitor. 
As you can see from the picture itself, this vector field has $\mathbf{e}_x$ component and also a $\mathbf{e}_z$ component (contrary to what your simplistic drawing suggests).
If you consider two of them and place them separated from one another (say 10 meters apart) then I'm sure you can imagine that the field lines are extremely complicated. Analytically carrying out such line integrals are of course next to impossible, but it's a matter of experiment  (and thus theory) that such fields are conservative, so the loop integral is always $0$.
A: Without a magnetic field existing, $\oint{\vec{E}\cdot d\ell=0}$ is always true.
Your mistake is that you don't consider the edge effect of a parallel plate capacitor. As the picture you can see,

when you approach the edge of the parallel plates, the horizontal component of the electric field cannot be neglected. Counting this effect in, you can have two different electric fields, and the law of a conservative electric field is still true.
A: For the electric field vectors to be normal as you have shown it, it must be that the plates of the capacitor are considered infinitely large.
This will cause the two plates you've shown separated in space to be essentially portions of this ideal big plate considered for deriving the formula.
You could try and keep a different charge density on different parts of this mathematical plate, but for the electrostatic case, it will eventually be redistributed such that the field inside the metal is zero.
Therefore, no contradiction.
A: Another way to see it is to try to plot the potential. The region between each pair of plates is fairly uniform, and one is much steeper than the other, but the field in the region in the middle joins them smoothly. The potential along any path in the plane goes up and down, but the ups have to equal the downs around any loop.
The electric field is the gradient of the potential, you can think of it as a bit like $d\phi/dl$ as you move along the line. That is, $E\cdot dl$ is the amount the surface rises or falls as you move a distance along the path. This has to equal zero around any loop, as you must end up at the same height as you started, but there is no contradiction with different parts of the line having different gradients. The problem is using the uniform-field approximation beyond its range of validity.

