Fourier transform of frequency division
A sum of two sine waves can be expressed equivalently as the product of two sine waves.  This is a phenomenon commonly referred to in musical circles as a beat frequency, which can be used to tune instruments by ear.  For your example, your input can be expressed as a 103.5kHz frequency multiplied by a 1.5Hz frequency.

I wouldn’t use the term “squaring” for what you are doing as much as “clipping the ever living hell out of”.  This would probably be expressed best using concepts from intermodulation distortion.  However, in your application, the square nature of the output of the waveform can be analyzed more simply anecdotally.  I would expect to see a square wave at the 103.5kHz with glitches at the 1.5Hz zero crossings.  But notably, it would look a lot like a 103.5kHz waveform, with some side band nonsense caused by glitches.

I do not believe that the flip-flop can be expressed using any common signal processing techniques, but again it can be analyzed anecdotally.  As you’ve hypothesized, I would expect the time between edges to double, because you’re skipping every other edge.  However, this is a grosse generalization.  It would be exactly true for a sinusoid, less true for the input you’ve proposed, and untrue for a generic input.  So for you’re example, mostly energy at 51.75kHz.

At this point, you still have a square wave, roughly at 51.75kHz, with some side band garbage cause by the glitches.  One could apply an arbititary amount of filtering to extract the sine wave at this frequency, resulting in the output waveform you described.

Something else to keep in mind.  A frequency transform has practical limitations.  A proper transform requires all time domain points of the input signal.  Using less than infinite samples is equivalent to windowing, which causes side lobe patterns about dominant frequencies, which is a topic unto itself.  Additionally, a signal sampled at discrete time intervals (like on a computer) has a limited bandwidth.  When you “clip the ever loving hell” out of a signal, you are widening the bandwidth of the input signal significantly.  If care isn’t taken to ensure that the discrete time sample rate is high enough, a significant amount of aliasing distortion will occur, hindering proper analysis.

Sorry for being a bit of a rant, but your question touches on a lot of points.  Cheers.