For an INFINITE parallel plate capacitor, the electric field has the same value everywhere between the 2 plates. An intuitive reason for that is: suppose you have a small test charge +q at a distance $x$ away from the +ve plate and a distance $d-x$ away from the -ve plate. The +ve plate will repel the charge and the -ve plate will attract it. Now if the charge is at a distance $y$ from the +ve plate $(y>x)$ then the the repulsive force due to the +ve plate will be weaker but the attractive force due to the -ve plate will be stronger such that the net force on the test charge will have the same value as when the charge was a distance $x$ away from the +ve plate. This is because the electric field lines due to an infinite plate are parallel to the normal vector to the plate.
I don't quite understand your second question. If you have a fixed potential difference between the two plates, then the electric field between the plates is given by $E=\frac{V}{d}$, where d is the distance between the 2 plates. From the expression, you can see that the electric field is NOT independent of the separation of the 2 plates.
Now suppose you keep the charge on both plates the same and separate them, such the the distance, $d$ increases. In that case, the electric field will remain unchanged. This might seem counter intuitive at first sight, this might seem like a violation of the conservation of energy. But, the extra energy required to keep the E field the same comes from the work that you are putting in to separate the attracting plates. The capacitance, $C$=$\frac{\epsilon A}{d}$, where A is the area of the plates, d is the separation of the plates and $\epsilon$ is the permittivity of the material between the plates. If $\epsilon$ and $A$ are constants, then, $C$ is inversely proportional to $d$.
The relation between the potential difference between the 2 plates, $V$ and the charge on one of the plates, $Q$ is given by $Q=CV=\frac{\epsilon A V}{d}=\epsilon A E$.
$\implies E=\frac{Q}{\epsilon A}$
Therefore, E is independent of $d$ if the charge on each plate is unchanged.
Electric field at any point is the force that a unit positive charge (1unit positive charge= 1Coulomb) would feel when placed at that particular point. This is a good way to decide the direction of the electric field lines.
Basically we define electric field in terms of the force a particle experiences at that point because that's the only thing we can measure. Field lines are just there to help us visualise. Yet they are a very important concept in physics. They stress the importance of the finite time taken for propagation of electromagnetic waves in more advanced physics.
Best Answer
"Does it have to do with the attraction or repulsion between unlike and like charges?" Yes.
The electric field always goes out from positive charges and in towards negative charges. The magnitude of the electric field of the +10e charge is independent of the test charge.
\begin{equation} E=\frac{Q}{4\pi\epsilon r^2} \end{equation}
However, the direction of the electric field changes near the vicinity of the test charges. If the test charge is positive then the electric field lines would not meet since the field lines of both the particles are coming out of them and are thus in opposite directions. But if the test charge was negative then the field lines for one particle would be going inside while for the other particle it would be going outside. Since the field lines are going in the same direction (from positive charge towards negative charge) they would meet.
This will happen even if charge placed is $+10e$ while the test charge is only $1e$. Most of the region would be dominated by the field lines of $10e$ however, at some point, close to the $1e$ charge, they will cancel each other out and will not meet- causing the charges to repel.