A buck regulator to convert -48 V to -12 V at 50 A is possible, but it would not be an easy project to start on.
A buck regulator works by initially connecting an inductor between the input voltage and the output voltage. The current in the inductor builds up, storing energy in the inductor. When the current gets too large, the power is cut off, so no current is taken from the supply and the current comes from the inductor. The current reduces in the inductor, and the inductor transfers its energy to the output.
For small buck converters, there is a Schottky diode to allow the current in the inductor to flow to the output when the power is turned off. For larger supplies, the power loss in the Schottky diode would be too much so there is an additional transistor, usually a MOSFET, to short out the Schottky diode when the power is off. For a 50 A converter, you would need that. The technique is called "Active rectification" if you want to look that up.
A buck converter is usually quite efficient, so to produce 50 A on average at -12 V would only need an average of around 15 A at -48 V. The current will be taken in pulses, so when current is being taken, 50A or more will flow from the battery, and nothing when the power is coming from the inductor. That will result in problems and you will need large capacitors on the input and output of the buck converter.
For 50 A output, it would be better to have several buck converters arranged so that the current flow is more even. With 4 converters, producing 50 A in total, at any one time you would have one converter consuming 12.5 A from the supply while energy is building up in its inductor, while the other three would each be producing 12.5 A as energy is taken from the other inductors.
That is called a "Multiphase buck regulator" and it is explained here:-
https://www.ti.com/content/dam/vide...p4/subassets/4.2-multi-phase-presentation.pdf
That technique is used in computers where the processor runs on very low voltages but takes high currents.