The MOSFET is rated to handle that current when sufficiently cooled. The hole in the tab at the top of the chip facilitates use of a machine screw to attach the MOSFET to a heatsink. A MOSFET with a higher current rating will also typically be capable of a low ON resistance comparative to its physical size, and the ON resistance actually achieved will depend on temperature and gate voltage. You can find charts in the datasheet to see exactly what the ON resistance should be at the gate voltage you intend, and use Watt's Law (\$P=I^2R\$) to figure out how much power the MOSFET will dissipate at the current you intend. Your heatsink must be capable of dissipating this amount of power while keeping the MOSFET at your intended temperature. There will be a temperature gradient right from the semiconductors in the MOSFET to the extremities of the heatsink, which will be rated by this, in degrees C per Watt.
In the case of your mosfet, installed and cooled as intended, this amounts to \$120A^2*0.0024 \Omega=34.56W\$ used by the mosfet to switch 120A at 30V. In reality, switching losses may contribute considerably more to this depending on what you're doing.
The wires coming from the device are intentionally kept short, and usually reasonably chunky by electronics standards. They also benefit from the cooling of the MOSFET.