If we increase frequency of light, then energy of photons will also increase. We can clearly see this from this equation,
$E=h\nu$ ; $\nu$ is the frequency of light
We know that electrons lose energy on their way out of the metal lattice due to collisions. Some electrons lose all the energy due to the collisions before coming out of the metal and thus they cannot leave the metal lattice. But now since the energy transferred to them by the photons is more, those can now make their way out of the atom despite all the collisions. So, considering these electrons which now manage to come out, number of photoelectrons should definitely increase on increasing the frequency of light.
If the number of photoelectrons have increased, then saturation current, (photocurrent at which electrons with zero kinetic energy are also able to leave the metal lattice) will increase. It is because it may so happen that the electron which previously lost all its energy before coming to the surface of the atom will come out of the metal lattice with zero kinetic energy.
Therefore, this photocurrent vs anode potential graph at two different frequencies is wrong. Saturation current of light with more frequency must be more since the number of electrons that reach the other plate is more. ( Saturation current is actually a measure of photoelectrons that reach the other plate).
What am I missing out?
Please don't give any explanation based on formulas.
To illustrate what I am saying better,
Consider an electron. In situation 1, it absorbed 3ev (numbers are randomly taken) energy from a particular photon. On its way out of the atom, it faced collisions. Say, to overcome those collisions, it needs to give 4ev energy. But since the electrons does not have that much energy, it will not be able to come out of the atom. In situation 2, say, the frequency of the light increases, and therefore more energy gets transferred to this electron, say 5ev. Now, on its way out of the atom, it will have enough energy to face all these collisions and come out. Therefore, it becomes a photoelectron.