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.
You are correct in your reasoning. However, that line does have a voltage equal to $0$. The trick comes in from the fact that $V=0$ doesn't mean $\vec{F}=\vec{0}$. The force depends on the change in the potential, and, as you noted, it is changing; the potential just happens to have a value of 0 along that line. Think of it like an inland hill, where the bottom of the hill happens to below sea level but the top of the hill is above sea level. Just because the altitude is $0$ in the middle of the hill doesn't mean a ball won't roll right past it all the way to the bottom.
Electric potential works the same way as height does in the context of gravity. The electric force will always accelerate positive charges toward lower and lower potentials. The reason force doesn't have to be $0$ when potential is $0$ is because potential can and does go negative. The potential is $0$ on that line, but it's negative left of the line, so a positive test charge will be accelerated to the left.
You're right in thinking that potential is all relative and that saying it has a potential of $0$ is arbitrary. The value of potential is irrelevant; only change in potential matters. A very convenient convention, though, is to define the potential at $r=\infty$ to be $0$. That's where the equation you have comes in, and when we use this "gauge", the potential on that line can be found to be $0$.
That line, since it has the same potential ($0$) along the whole thing, is an equipotential line. The point of the equipotential line is that if we take a test charge, it doesn't require work to move it along the line, i.e. vertically in the picture. You are correct that if the particle were to move horizontally, then there would be work done on it by the field. But it's not moving horizontally along the line. An equipotential line is always perpendicular to the electric field lines for this reason, which the line in the picture clearly is. Work is defined as $W=\vec{F}\dot{}\vec{d}$, and since $\vec{F}\perp{}\vec{d}$ along an equipotential, $W=0$.
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Consider two equal and opposite charges ($+q$ & $-q$) in space separated by a distance $2r$. An uniform electric field would exist between both acting from $+q$ to $-q$. The first thing is, Electric potential is a scalar quantity whereas Electric field is a vector..! In other words, Electric field is a measure of how the electric potential changes quickly with distance (gradient or the first derivative).
The electric potential at a distance $r$ from $+q$ would be $V_1=\frac{kq}r$
Now, the electric potential at a distance $r$ from $-q$ is $V_2=-\frac{kq}{r}$
The net (effective) potential at midpoint ($r$) is $V=V_1+V_2=0$
In case of Electric field, it is non-zero. Because, we would specify the direction only...
Regarding your case, A test (point) charge not necessarily positive. It's just to indicate the existence of an electric field. In presence of a charge, the test charge would experience a force. That's all :-)