The actual value of the potential is not meaningful, what is meaningful is the gradient: the field on a particle is given by $\vec{E} = -\vec{\nabla} V$. Equivalently, what is meaningful is only the potential difference between 2 points.
Notice that one can define a new potential differing from the old by an additive constant: $V' = V + c$ and the force will still be the same. So we have a redundancy in the system when describing potential energy, this is a gauge degree of freedom. But nevermind the jargon for now.
In your case, the test charge (positive)'s derivative of its potential energy will give rise to a force towards the negative charge, and so it moves towards the negative charge. Conservation of energy holds, and the gain in KE must be compensated by a loss (i.e. difference) of PE.
The potential being $0$ at the midpoint is a consequence of choosing $c$ such that the potential at the point $r\to \infty$ has $V \to 0 $, but one can equally well define the potential at $\infty$ to be $1$ GeV, for example, and nothing physical would change.
Yes, TO ACTUALLY MAKE IT MOVE a net force must be greater, however let's say I provide a force infinitesimal greater, it will have some velocity. And then moving along that distance I will have done X amount of work against gravity.
The way you should think of it, is that IF an object moves through a distance, the Electric field is going to either be doing positive or negative work per unit charge.
Meaning if lets say in moving through a distance A to B the field does -W work per unit charge. The total amount of energy I need to provide to overcome that negative work, will be "W" aka the negative of the work done by the field. As W+(-W) = 0 so if I give W work over a distance, then the field will equally "Steal" that work from me, causing no net gain of KE by the object
Its all about measuring how much work is either done or lost on an object by the field IF it were to move along a certain path. The negative of this is the total amount of energy that "i" put in, to overcome this
Thought experiment:
Given an object moves vertically with an initial velocity V, as the object moves gravity will do negative work on it, such that at a height "H" the total amount of negative work done by gravity is equal to the objects kinetic energy, aka the object STOPS at height H
now imagine that now, the object is at the ground with the same initial velocity. But this time, I am applying an EQUAL BUT OPPOSITE force to gravity. The net force is zero, so the object will continue to move upwards at a constant velocity. When the object reaches a certain distance upwards "A", I will then STOP that applied force, clearly now the object is at a height "A" and will then continue to move upwards by the same amount as the first scenario and then stop.
Meaning that I have now applied a force -F over a distance A but this time the object has now has a maximum height A+h
Aka, A meters higher. Clearly the work that I have put in, has caused the object to move A meters heigher, than the previous scenario.
Aka, that is the total amount of work I have to do in order to overcome the force of gravity through a distance A.
More mathematically.
Consider the equation
$1/2 m v^2 + \int_{A}^{B} F \cdot dr = 0$
Aka, an object with an initial Ke, moves through the distance a to b, in the presence of a force field, such that when it reaches B, it stops. Well how much energy do I need to give the object such that in the presence of F, it stops when it moves through a to b?
Clearly
$1/2 m v^2 = - \int_{A}^{B} F \cdot dr $
this is exactly the same as if I apply the work to the object over the distance, instead of at the very start in the form of Ke
Best Answer
If the charge comes from a point of $V_3=0$ and gains kinetic energy, then it is because the value of potential energy at the new point is lower. Then potential energy can be converted into kinetic energy. Don't worry about the actual potential energy values - only the differences in value matter.
Think of having a box on the floor. You might say that there is zero (gravitational) potential energy associated with it. But that is only because you chose to consider the floor as the reference.
The values don't matter. What matters is only that some values are smaller than others, because the box will always want to fall towards lower values. It will fall from the high shelf to the floor at zero potential energy, and it will fall from the floor at zero to the hole at negative potential energy. It always wants to move towards lower values - the actual value doesn't matter.
You are free to pick whichever point you want as the zero-value reference. It doesn't matter, only the difference between points matters.
The same is the case for electrical potential energies. You could place a positive charge at the shown equipotential line and say that zero (electrical) potential energy is stored. Then surely, the charge will want to move towards the neighbour locations where the potential energy stored is less than zero. That somebody chose the potential energy values at this particular equipotential line to be zero, doesn't matter. It could have been anything else.
This tendency to move towards lower values of potential energy is what the field lines show. At all points on the equipotential line, there are field lines showing the direction that the charge wants to move along.
In general, you should forget about the actual values of potential energies and only care about the differences in the value between points. This is why voltage is the main parameter in these cases; voltage is the difference in electrical potential between two points. Just pick whichever reference that makes it easier to work with in your specific scenarios.