Your 1st equation can be interpreted as saying that the flux through the section is equal to $\frac{1}{\epsilon_0}$ times the charge $Q$ contained within a cone whose base is the section and whose vertex is the centre of the sphere.
This result can be obtained directly using Gauss' Law. The electric field inside a uniformly charged sphere is radial. So the flux across the slanting curved face of the cone is zero, because this face is radial. The only flux out of the cone is across its base. The electric field across this base varies in magnitude and direction. Nevertheless, the total flux across it equals the charge $Q$ enclosed by the cone divided by $\epsilon_0$.
Your 2nd equation is derived using Gauss' Law and gives the total flux through the surface of the cylinder.
The similarity between the formulas for $\phi_1$ and $\phi_2$ is entirely due to geometry. Gauss' Law says that total flux through a surface of any shape containing uniform charge density $\rho$ is $$\phi=\frac{\rho V}{\epsilon_0}$$ The only difference for each shape is the volume $V$.
The volumes of cone and cylinder depend in the same way on base area $\pi a^2=\pi (R^2−r_0^2)$ and height $r_0$. They are both of the form $k r_0 \pi a^2$. For a cone $k=\frac13$ while for a cylinder $k=1$.
You can see by looking at extremes that your conjecture (that the flux through the flat ends of a cylinder is $\frac13$ of the total flux) must be false. For a short fat cylinder ($r_0 \ll a$) almost all of the flux will be through the flat ends. For a long thin cylinder ($r_0 \gg a$) almost none of the flux will be through the flat ends.
