Why is the loaded Q of a critically coupled two-resonator circuit approximately equal to 0.707 times the loaded Q of one of its resonators?
This is just an approximation.  If you look at it in terms of voltage 0.707 is 3dB drop or 1/2 power.  Adding a second coupler will increase the bandwidth which will cut the Q by 0.707 as compared to a single resonator.  If you want to compute this just compute the 3db points for a single resonator vs. a critically coupled set.  You can do this by analyzing the transfer functions for each circuit and solving for the 0.707 points.

You'll find the low frequency side of this network to move at 18db/octave and 6db/octave on the high end due to it's poles and zeros.  This lack of symmetry in the system make the change in Q (i.e. bandwidth) an approximation.