Well, to put it shortly, it's just because $L^2$ and $L_z$ are two observables that have no $r$ dependence. Since the kinetic terms of $H$ can be written as functions of $r$ and $L^2$, and the potential depends only on $r$, it is clear that the Hamiltonian commutes with the angular momenta. As such, it makes sense to write down $Y$ in terms of these two.
But if we'd like to go a little deeper, it is a bit related to representation theory. The fact that $L^2$ and $L_z$ form a complete description of vector rotations (they represent the Lie algebra $\mathfrak{so}(3)$ of the Lie group $\mathrm{SO}(3)$) and therefore are the most natural way to express the angular dependence of a rotation-invariant space in terms of orthonormalizable functions.
After reading @ZeroTheHero 's comment, i figured it would be instructive to provide some intuition as to why angular momentum does not depend on radius. Remember the classical definition
$$\mathbf L=\mathbf r\times\mathbf p$$
Let's then give and estimate of $L$. Suppose an atom or something has some characteristic radius $R$, and the electron moves with some characteristic momentum $p$. Then we estimate
$$L\sim Rp$$ But the de Broigle relation gives us an estimate of momentum in terms of the characteristic length
$$p\sim\frac{h}{R}$$
Where $h$ is Planck's constant. As such, out estimate for angular momentum amounts to
$$L\sim h,$$
Which has no explicit dependence on characteristic lengths of the system. As such it is reasonable to say it has no $r$ dependence, even though I've not given a rigorous proof.