And again why put a 2A fuse at F3.
The only real reason is to lower the resistance of the fuse with another fraction of an ohm, each fraction of an ohm, wen drawing 1A trough it (that is 400 Ma drawn by the PI itself, one USB port drawing 100mA, and another one drawing 500mA) would lower what they all see by a fraction of a volt, and that fraction of a volt can be relevant!
For example if you draw one amp through a PSU that is able to deliver exactly 5.0 volt at that amperage, but your USB power cable has a realistic resistance of say 0.1 Ohm, your PI at it mini USB port would receive 0.1 volt less, so 4.9 volt, now if the F3 fuse was an 1A one (a 0.2 Ohm fuse) then it would drop 0.2 Volt, and the PI would receive 4.7 volt, and would fail (minimum is 4.75V) if the fuse would be a 2A one and thus only 0.1 Ohm it would drop only 0.1 volt, and the PI would get to work with 4.8 Volt. Obviously this is just an example, but I just wanted to explain that indeed just 0.1 ohm more or less polyfuse resistance can mean a working or a failing system.
Just a few "rule of the thumb" figures for normal copper thickness of 0.17mm (half ounce per square feet).
A trace of 5 cm long and 0.5mm wide would be about 0.1 Ohm.
See : http://circuitcalculator.com/wordpress/ ... alculator/
So if that is more or less similar to the trace from the PI's input connector to the USB ports, (but I think the trace would be wider than half a mm) then running 1A through them would result in a 0.1 Volt drop. It would also mean that the dissipation (energy turned into heat) in the total length of the trace would be P=U*I = 0.1 x 1 = 0.1 Watt, which is such a small number that it would hardly warm up the trace. If the current would be double that (2A) the energy dissipated in the trace would be four times that, or 0.4 Watt, still not really significant.
By the time the current would reach levels which would be threatening to the trace, the fuse would have been long blown, and unless the "PSU" is a 200 Watt PC PSU the 5V would have collapsed completely. I simply see no valid scenario where the traces would burn out, unless you are able to keep up 5V while sending 10A trough the trace, which is absurd when using a normal PSU.