What is radial probability density?

Those factors usually come out when separating the wave-function for a 3-dimensional problem (like the Hydrogen atom) in its radial and angular parts: $$ \psi(r, \theta, \phi) = R(r) Y(\theta, \phi). $$

If you are then interested in the probability of finding the particle (or whatever you are studying) at a distance from the origin between $r$ and $r+dr$ around the solid angle characterized by the angles $\theta$ and $\phi$, you have to integrate $\lvert\psi\rvert^2$ between $r$ and $r+dr$. In spherical coordinates this reads

$$ \lvert\psi(r,\theta,\phi)\rvert^2 \hspace{-8pt}\underbrace{d^3r}_{r^2\sin\theta drd\theta d\phi} \hspace{-10pt} = r^2R(r)^2 \lvert Y(\theta,\phi) \rvert^2 \,\,\sin(\theta) \,\,dr d\theta d\phi $$ hence the term $r^2 R^2(r)$. If the problem at hand is spherical, it can be more interesting to ask what is the probability of the particle being found in the spherical shell between the radii $r$ and $r+dr$, regardless of the solid angle. This amounts to integrate the above in the whole solid angle, which if there is no angular variation of $Y$ (that is, if $\psi(r,\theta,\phi)=\psi(r)$), gives $4\pi$:

$$ \operatorname{P}(r_0 \le r \le r_0 + dr) = 4\pi r_0^2 R^2(r_0) dr $$