What do they mean when they say that it does not require any work to move a charge from one point to another in an equipotential surface? In the textbook it says that no work is required to move a charge from one point to another on an equipotential surface. Do they mean work by the electric field or work by anything? Because clearly the object cant just magically move sideways with nothing.
 A: "it takes no work" in the same sense it takes no work to move an object on a perfectly frictionless, flat surface. It is true and theory, but moving an object requires accelerating it at least a little bit, which requires some work, as you point out.
A: It takes work to transfer kinetic energy into the charged object and get it moving, sure. But if the object was already moving, it wouldn't lose any energy by moving along the equipotential. Furthermore, you could decrease the work needed to stop and start by just moving slower, with no work needed in the limit of infinite time taken. The work of starting and of stopping can cancel each other out so that all your energy can be recovered in an ideal system. So if work is needed to move a charge along an equipotential, it isn't because of the electric field. 
A: If the particle is on an equipotential surface, then that means there is no force from the electrostatic field on your charge, while moving along that surface.  If there are no external forces, that means that locally momentum in that direction is conserved, just consider Newton's laws:
$$ \frac{\text{d}\boldsymbol{p}}{\text{d}t} = \boldsymbol{F} = 0 $$
However in moving your particle, you would change the velocity, so we require
$$\Delta \boldsymbol{p} = m_q \Delta \boldsymbol{v} \neq 0$$
Hence, to move your particle along an equipotential surface, you don't need to do any work, but you do need supply a change to the particle's momentum.  Hence the particle can't spontaneously change its trajectory while on an equipotential surface.  The issue here is when changing the momentum, in general you will change the energy too, and end up doing some work with respect to whatever external field you're using to push or pull your charge around.
