No1- The square wells are approximations, and in general don't show up in reality (reality doesn't like discontinuities).
No2- as an object V=V_{0} you can find that the force (being the gradient of the potential) would be something like F= 0. That is something normal, since a constant potential wouldn't make any difference (V=V_{0} or V=0 are exactly the same).
So the difference will come from the point of discontinuity, where the potential jumps to a given number...
A step potential for example would have to be like that:
V= V_{0} Θ(\frac{a}{2}-|x|)
where Θ(x-x_{0}) is the Heaviside step function, being +1 when x>x_{0} and 0 for x<x_{0}
So in fact, the gradient of it, will give you the delta functions at the points x=\pm \frac{a}{2}...
So it's like the particles withing the step potential will "feel" infinite forces when trying to leave it keeping them "inside"... (because we are speaking of a well)
Of course, once you get into QM you must start stop thinking about the "forces" and get used in reading potentials... That's why my calculations may be wrong, but the general idea doesn't change... If someone would sit and evaluate the -gradV, he'd eventually find a better result, which in QM is unnecessary
ALSO- the step potential in nuclear physics is unphysical- it's only a good toy and gives some approximate results- however it lacks other things (for example coulomb potential, centrifugal potential etc)