When an open-drain output asserts low, where does the line capacitance discharge?
The charge on \$C_{RETRY}\$ goes to ground through the \$\overline{FAULT}\$ pin.

Although there is no external resistance, the internal transistor that switches that pin to ground will not be a short circuit, but will have a lowish current limit. As you are being told by the manufacturer to make that connection for that purpose, you can assume it's OK to do so.

The absolute max current specification on that line specifies a continuous current. This puts a lower limit on the size of \$R_{FAULT}\$.

The energy on \$C_{RETRY}\$ ends up being dissipated as heat in the pin driver transistor. This is a small and limited amount of energy which has to be distributed in the thermal mass of the transistor, unlike the energy due to a continuous current which is unlimited, so has to be dissipated.