I never noticed that, and didn't think of it for myself either - shame on me! I hadn't actually read much of the text, and wrongly assumed the paralleled mosfets were to increase current capability, when in fact they are to decrease the combined "on" resistance. Oops... With a 0.1 ohm sense resistor it would need a rating of 1000W to be able to pass 100A. Well that's never going to happen. I am a dumass...
Anyway, no need to worry about safety aspect of something capable of handling a large current, apart from its ability to get hot. It's not the excessive ability of a device to cope with extremes you have to worry about, it's the extremes themselves, and inadequately rated components.
Ok, quick calculation. The article talks about 7 amps because the 0.1 ohm resistor is rated at 5W, so I'll use that. The mosfet in use has an Rds(on) of 0.022 ohms. So with 2 in parallel, that's 0.011, put 7 amps through that, you get 7x0.011=0.077 volts. With only 1 transistor, you get 7x0.022=0.154 volts, and only 1.078 watts of power dissipation (and only 0.539 for the 2 transistors in parallel). So the question becomes, under what circumstances are you going to want to load something where the voltage is that low? The only thing I can think of is foldback current limiting of a PSU - ok so it might be handy if you design psu's... But even so, you have to remember there is a 0.1 ohm resistor in series anyway, which at 7 amps is going to drop 0.7 volts, effectively swamping the voltage dropped by the mosfets.
Supposing you want to test a 12 volt PSU, rated at 3.5amps, with a 3 amp load. That gives you 36 watts which your load has to dissipate, and it has to set it's mosfets to have a resistance of 4 ohms (well ok, 3.9 because you have the sense resistor in series) - way above the low combined Rds(on) you get with this design. But suppose you crank the load so it now draws 4 amps. If the PSU has foldback current limiting, the voltage will drop right down, and depending the design it may sustain that amount of current or it may pop a fuse, but what should happen is that the voltage drops down enough that the constant current load reaches it's minimum resistance, so only 3.5 amps is delivered. Foldback is conceived with the idea of constant resistance loads, where if the voltage is dropped down, at some point the current will fall within the design limit of the supply.
Hope this clears some things up a bit