What is the entropy change of the universe for a rock if it falls from a height into a lake? The rock and the lake are at the same temperature

The lake absorbs part of heat through the friction between the rock and the water's mulecules. At the same time, the internal energy of rock is changed accordingly. Therefore, this process (heating up the lake) is irreversible. So the total entropy increases.

The rock dropping all by itself (without friction) is a completely classical (non-thermodynamic) process, so the stone alone dropping causes no entropy change. The change in entropy comes from excatly the heat absorption process you describe.

At first, the stone has some potential energy. During falling, this is converted into kinetic energy. By hitting the water, the kinetic energy of the rock is then converted to heat and absorbed by the whole system of lake + stone. This heat absorption is what causes the entropy change. For constant $T$, we have $$Delta S = frac{Delta Q_text{rev}}{T}$$ and in our case $Delta Q_text{rev} = mgh$.

The first step in determining the entropy change is to apply the first law of thermodynamics to establish the final equilibrium state. In this case, the potential energy of the rock is converted to internal energy of the rock plus surroundings (rest of universe): $$Delta U+Delta (PE)=Delta U-mgh=0$$where U is the internal energy of the system, which, in this case, is the combination of roc plus universe. So, $$Delta U=mgh$$

This change in internal energy is virtually all in the surroundings, since the final temperatures of the rock and surroundings are the same, and since the mass of the surroundings is so much larger than the rock. Treating the surroundings as an ideal infinite reservoir, we have that $$Delta S=frac{Q_{rev}}{T}=frac{Delta U}{T}=frac{mgh}{T}$$where $Q_{rev}$ is the amount of heat that must be transferred to the surroundings in an alternate reversible path (in which it experiences the same internal energy change as the actual irreversible process).