Your question doesn't have an answer because it isn't possible to split energy into a potential part and a kinetic part, or at least not in an observer independant way.
The Schwarzschild metric is time independant and spherically symmetric, and these symmetries mean there are two conserved quantities that you can think of as total energy, $E$, and angular momentum, $L$. We can use these symmetries to derive an equation for the dependance of the $r$ coordinate on proper time:
$$ \left(\frac{dr}{d\tau}\right)^2 = E^2 - \left(1 - \frac{2M}{r}\right) \left(1 + \frac{L^2}{r^2}\right) $$
and this looks like an equation of motion with an effective potential $V_{eff}$ given by:
$$ V_{eff}^2 = \left(1 - \frac{2M}{r}\right) \left(1 + \frac{L^2}{r^2}\right) $$
This is the equation you give (you've subtracted off the $1$, but that makes no difference to the working), but although it has the form of an effective potential this isn't a potential energy.
A graph of $V_{eff}^2$ against $r$ looks like:
![V_eff^2](https://cdn.statically.io/img/i.sstatic.net/FvxBS.gif)
and as you say, we get two circular orbits at the turning points (marked with arrows).
If you wished you could solve the equation for $V_{eff}$ at these points, but the result is not a potential energy in any useful sensed of the word.
You don't say what texts you're using to learn GR. If you're mainly driven by curiousity I strongly recommend A first course in general relativity by Schutz. The material above is dealt with in Section 11.1.