The simulation shows the equipotentials for a uniform field. In which direction is the field?
Field lines are always perpendicular to equipotential surfaces, so the field is vertical. The electric field points in the direction of decreasing potential, so that's down.
Let's say a +q test charge is moved horizontally a distance r. What is the change in potential experienced by this charge?
In this case moving horizontally means moving along an equipotential, so the potential does not change.
Now the +q test charge is moved vertically a distance r. What is the change in potential experienced by this charge now?
The charge moved from the V = 4 volt line to the V = +8 volt line, so the change in potential is +12 volts. Note that the change in potential matters much more than the actual value of the potential.
How would your answer change if the charge had been a negative charge, q, instead?
The change in potential is unchanged, although the change in potential energy in the two cases would flip sign.
Consider two cases. In case 2 the equipotential lines are twice as far apart as they are in case 1. In this simulation each line differs from neighboring lines by 2 volts. If a +q test charge is moved vertically a distance r in both cases, in which case does it experience the largest change in potential energy?
Case 1, because the change in potential is twice as large.
In which case is the electric field strongest?
Case 1  Field is a measure of how quickly potential changes. The closer the equipotentials the stronger the field.
Summary: in a uniform electric field the change in potential is:
DV  = 

=  E • Δr 