The relation between the potential $\Phi$ and the electric field $\vec{E}$ is:
$$ \vec{E} = - \nabla \Phi $$
So the field is not strong when the value of the potential is high but when the local change in the potential is high. Vice versa, if the electric field is zero, the potential might still be at a very high value - it just does not change at that point.
It is a classical confusion for most people learning electrodynamics, but e.g. a potential of zero does not mean that the field there vanishes and vice versa a field of zero does not imply anything about the value of the potential.
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.
- Lift the box to a shelf, and there is positive potential energy stored.
- Put it in a hole in the floor, and there is negative potential energy stored.
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.
Best Answer
Here are the two arrangements of charges that you have mentioned shown as potential, $V$, against position along line joining charges, $x$, graphs - the one dimensional case $E_{\rm x} = - \frac{dV}{dx}$
If you added $20$ to all the potential values does that change the shape of the graphs?
The answer is "No" and this then means that the gradients of the graphs at positions $A$ and $B$ do not change.
When you find the gradient the important parameter is the difference between the potentials and so it does not matter if you have $[3-5]$ or $[(3+20) -(5+20)]$ the result is still $-2$.
As expected at position $A$ the gradient (= - electric field) is zero and the potential can be any value you choose it to be and at position $B$ the gradient (= - electric field) is negative which shows the electric field to be positive ie in the positive x-direction towards the negative charge.