Yes the additional positive charge to one plate increases electric field behind the other one.
To see this let's look at general Gauss's law:
$$\oint_{S}\vec{E} \,d\vec{A}=Q/\epsilon_0$$
The $S$ is the Gaussian surface which we are free to choose therefore it could include both plates even if the surface goes to infinity with both plates.
The charge Q however could be written as charge density $\sigma$ times surface area or in general:
$$Q = \int_A \sigma \, dA $$
where $\sigma$ could be $\sigma_1$, $\sigma_2$ or $\sigma_1 + \sigma_2$.
Now we are ready to use symmetry of configuration. Lets assume that field is homogeneous at the middle of of both infinities and rewrite the left side of Gauss law as:
$$\oint_{S} \vec{E}\, d\vec{A} = \oint_{S_{in}} \vec{E}\, d\vec{A} + \oint_{S_{out}} \vec{E}\, d\vec{A} $$
The right side with the same division:
$$\int_A = \int_{in} +\int_{out}$$
Now we are ready for the main part:
$$\oint_{S_{in}} \vec{E}\, d\vec{A} + \oint_{S_{out}} \vec{E}\, d\vec{A}
=\int_{in}\sigma \, dA + \int_{out}\sigma \, dA$$
And the Gauss law says that charges which is outside from closed surface does not affect the flux or $\vec{E} = \vec{E}_{in} + \vec{E}_{out}$ and the $\oint_{out} \vec{E}_{in} \, dA = 0 $ which makes to satisfy:
$$\oint_{S_{out}} \vec{E}\, d\vec{A}
= \int_{out}\sigma \, dA$$
And the only thing which is left is always satisfied:
$$\oint_{S_{in}} \vec{E}\, d\vec{A}
= \int_{in}\sigma \, dA$$
which for homogeneous field means:
$E_{top} + E_{bottom} = \sigma / \epsilon_0$
where the signs for field is positive if it is directed out of surface.
Coulomb's law is indeed a special case between two point charges. to find the force between a point charge and a plate, you would have to integrate the equation over the plate surface to calculate the contributions from all infinitesimal charge elements.
It's more practical to figure out what the electric field is, and then use $\vec{F}_e = q\vec{E}$ to find if there is a force applied to your test charge.
The electric field from an infinite plate with uniform surface charge density is given by $\vec{E} = \frac{\sigma}{2\epsilon_0}\vec{a}_n$ where the $\vec{a}_n$ vector is pointing away from the plate. Therefore, if you have two parallel plates (sufficiently large compared to the distance between them to be considered infinite) with the same $\sigma$ charge density, the electric field between the two will be null, and no force will be exerted on the test charge.
Best Answer
Electric field is a vector. It can point left, right, up, down, forward or backward. In your example it will point from the positively charged plate to the negatively charged plate. Whether you consider that positive or negative depends entirely on your choice of what direction to call "positive" and how you arrange the plates.
If you say that electric fields pointing to the left are positive and ones pointing to the right are negative, and then arrange your capacitor with the positively charged plate on the right and negatively charged plate on the left, then the field will be "positive". But if you turn the capacitor around and put the positively charged plate on the left and negatively charged plate on the right, then the field will be "negative".