Proper behavior for USB and +5V
I don't understand what you would use the USB's 5 V for. You have a 7805 to get 5 V from your 12 V, which you'll need anyway, so the 5 V will always be available, and with enough current without needing that 100 mA from the USB port. So I would disconnect the latter by cutting the trace at the connector, then you're sure it can't cause any conflicts.  

The reason why you should never directly connect outputs from voltage regulators is that the world isn't perfect. If 5 V would be 5.000000 V you wouldn't have a problem. But regulators have a tolerance on their output voltage, often 5 %, so that 5 V may as well be 5.25 V. So most likely the two voltages aren't equal.  

The second part of the problem is that voltage regulators have a very low output impedance, like a few 10s of m&Omega;, for an ideal regulator it would be zero. So a voltage difference combined with a very low resistance gives you a high current.

If you want to combine outputs from 2 regulators use Schottky diodes. They will give you a voltage drop of a couple of 100 mV, which may be acceptable. If you want you can add the same diode in series with the 7805's ground pin. Then you first lift the output to 5.3 V and a 0.3 V from the series diode will give you 5 V out. With a series diode the USB output will be lower, then the highest voltage has the priority.