If F1 and K1 are through hole parts, you could put two tracks in parallel, one on each side of the board. Each track would be roughly 10 - 15 mm wide.
Another option, Don't draw the connection as a track but as a polygon or fill. Most of the current will still follow close to the shortest path, but the added area will help to dissipate heat.
Also, be aware that for such short distances, the current crowding to get to the individual pins of your components may dominate the overall resistance of the connection. Using larger holes (requiring components with larger leads) will help with this.
Edit
First, I'm not sure how this plays in, but this ampacity table claims a 0.25 inch trace in 2 oz copper is sufficient to carry 24.5 A with 30 C temperature rise. This is about 1/4 of the number you are coming up with (25 mm). Unfortunately the source is not clear about what assumptions went into their calculations.
Second, the math you added in your edit looks fine.
As you see, increasing the copper area allows you to dissipate more heat. There's no reason you should only increase the area by increasing the trace width. You can simply create a large copper area of whatever shape is convenient the closer to a simple square or circle, the better), so long as it connects your input and output, and the heat will spread readily to allow dissipation over the whole area (with, of course, some concentration in the region where the heat is actually being generated).
Third, no back of the envelope calculation is likely to be especially accurate for these kinds of thermal estimates. If you really want to get a good estimate, you should look into a thermal analysis tool like FloTherm or Ansys IcePak. These tools will both improve the ability to estimate convection effects accurately, and take into account "edge effects" that come into play because your "trace length" is so much shorter than your "trace width".
I'm curious how you derived your power dissipation numbers. Looking at the data sheet it looks like 10ams 200 mW (12 degree temperature rise), 30 amps, 2.5W with a 90 degree temperature rise (given the Rthja of 40 degrees/W which seems to be true even if you have 6 cm^2 of PCB area).
That said, if you want to pull a lot of heat out of your FETs you can have a .250" plated through hole drilled under them and then use a copper slug which extends up through the hole and contacts the back of the package. you could also glue a heat sink to the top but it is not as effective trying to conduct through the case.
To your layout questions, it looks like a 6mil trace for all of the source leads. That would be a poor choice at 30A, by comparison look inside a 30A fuse :-) What it means is that you'll get some warming on that trace. What ever trace width you pick, do the calculation at your chosen copper level and use current squared x resistance to compute how many watts that trace will dissipate.
You don't need a all the vias you've got on the pad. 5 would be sufficient to thermally connect top to bottom. I've seen people just use one, but you rely heavily on the plate though of the hole in that case.
Best Answer
They are using power planes, maybe even with double-thickness copper layers too (70 microns / 2oz, rather than the usual 35 micron / 1oz Cu thickness), so the power path to the outputs is not going to be high resistance, and therefore not waste much power as heat loss, and therefore will not get hot and explode and otherwise be unable to "handle" the amount of current shown.
120A for 8 outputs is not too bad, each contact (which looks like just exposed mask on copper planes with holes for thermal relief) only needs to handle about 15A each which is entirely possible with nice fat wires going out to the rest of the set-up.
I would be worried though, first about the size of the battery needed, and second about your safety, if 120A loading was a common thing in your system.
You must also understand that using a trace width calculator makes certain assumptions, for example what the ambient temperature is, what the acceptable rise in temperature of the traces is, and in some cases how long the trace is, and finally the thickness of the trace which is usually measured in micrometers or ounces/square foot (oz). If you allow up to 30-40 degrees Celsius increase in temperature of the traces (which is usually fine for power related systems, which are often rated to 120-150 degrees) you can get away with very thin (2-3mm) tracks for carrying 15A, let alone how much you can do with a whole power plane (could be modeled as a 30-40mm wide trace!).