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 $4\pi$:
$$ \operatorname{P}(r_0 \le r \le r_0 + dr) = 4\pi r_0^2 R^2(r_0) dr $$
Best Answer
You are confused between the radial part of the eigenfunctions and the radial probability density.
For the 1s level of a hydrogen atom, the eigenfunction is $$\psi(r,\theta, \phi) = \frac{1}{2\pi}a_0^{-3/2} \exp(-r/a_0) $$ and there is no angular dependence.
But when you want to work out a probability density $P(r)$ for the electron to be found between $r$ and $r +dr$, then you need to consider an integral of the square of the modulus of the eigenfunction over the volume enclosed by the spherical shell between $r$ and $r+dr$ and this volume is $4\pi r^2\ dr$.
In other words, the (unnormalised) probability density as a function of radius $$ P(r) = 4\pi r^2 \psi(r) \psi^{*}(r)$$.
So whilst $\psi(r)$ peaks at the origin, $P(r)$ is zero at the origin. To work out at what radius the electron is most likely to be you look for a maximum in $P(r)$.