How can effective potential be used-angular momentum I do not understand effective potential, I have seen the derivation and I have seen the explanation that uses centripetal force.
My problem: Centripetal force is provided by the gravitational pull and therefore centripetal potential should already be in gravitational potential.
Problem 2: Derivations of effective potential get to this stage 
total energy = radial KE + term with angular momentum + potential energy. At this stage it seems to me that the term with angular momentum  is part of total KE, which makes mathematical sense, the the next magic step is to shove it in with potential and this makes no sense to me.
 A: You are correct that the gravitational potential accounts for the centripetal acceleration. However since angular momentum is conserved as $r$ decreases more energy goes into rotational kinetic energy. If we want to reduce everything to just the radial coordinate we no longer 'see' this rotational motion but the energy still is conserved, so we need to give up radial kinetic energy if we decrease r. This acts just like a potential.
As far as your second question, take the Lagrangian for a 2D particle in a central potential,
$$L = T-U=\frac{1}{2}m \dot{r}^2 +\frac{1}{2}m r^2 \dot{\theta}^2- U(r)$$
Then from the Euler-Lagrange equation for $r$
$$\frac{d}{dt}\frac{\partial L}{\partial \dot{r}} - \frac{\partial L}{\partial {r}}=m\ddot{r}+U'(r)-mr\dot{\theta}^2=0$$
From the equation for $\theta$ (angular momentum conservation)
$$-mr\dot{\theta}^2=-\frac{L^2}{mr^3}=\frac{\partial}{\partial r}\frac{L^2}{2mr^2}, $$
so our equation of motion is
$$m\ddot{r}=-\frac{\partial}{\partial r}\left(U(r)+\frac{L^2}{2mr^2}\right)=0$$
so the effective potential does appear with the correct sign, even though if you first plug in the constant $L$ into the Lagrangian before you find the $r$ equation of motion (which is not valid), it appears to have the opposite sign.
