Why does this constant-Q transform "lose octaves"?
The lower frequency atoms are longer in time (or of a length roughly proportional to the period of the frequency, give or take 2X).  So less of these lower frequency atoms are needed per second to cover that time period.  And more of the shorter atoms (for shorter wavelength frequencies) are required for about the same coverage (or with roughly the same percentage overlap).

And to maintain a nearly constant Q, the length of a bandpass FIR filter ends up roughly proportional to some constant multiple of the wavelength of the center bandpass frequency of that filter.