The bridge is still a bit of a black box in my mind (I get what it is does to the signals, I just don't get how it does it!) ...
There is a simplistic first-principles method to understand what the bridge does, and a more complex and accurate way to understand what the bridge does.
Look at the image of a bridge circuit below. This is exactly the same as the Vox circuit I described, as far as the first split-load inverter and the RC components after it.
The tube itself is replaced by the a.c. voltage generator in the diagram (the squiggle inside the circle). Call the upper output of the generator "the plate output," and "the lower the cathode output." Z1-Z4 each take the place of a Resistor-Cap pair (a couple of those are series-resistor-cap, a couple are parallel-resistor-cap). Where "Null" is shown are the 2 outputs which run to the differential amp stage I described before.
Not shown are 2 parallel resistor-cap pairs which run from the 2 outputs of the bridge to ground. They are both parallel-resistor-cap pairs. Forget they exist for now.
Bridge Operation:Look at the bridge as having 2 parallel paths from plate output to cathode output. Each of those paths is made up of an RC network instead of Z as shown in the simplified diagram.
In guitar amps, we typically think "caps roll off lows" because that is often the sonic impression that results. In reality, the cap has a changing reactance (depending on applied frequency) so that it forms frequency-dependent voltage divider with the resistor; in other words, the attenuation of the circuit changes with frequency.
Look at the upper line of the Filter Plot below, which is Amplitude vs. Frequency. The RC filter is a simple low-pass filter (a series resistor and a cap-to-ground), so attenuation is zero at a low frequency then gradually more as frequency rises. That's easy enough to grasp and fits with the basic understading of how caps work in an amp.
But reactive components store and release energy into a circuit, so the phase of applied voltage and applied current is not the same, as it would be in a resistive circuit/divider. So the phase of the voltage at the output of the filter also varies with frequency (if you need to know the exact phase at one frequency, you will need to use vector algebra to determine it). So the lower graph on the Filter Plot is Phase vs. Frequency. At zero attenuation, phase shift is zero. Phase shift approaches 90 degrees as the circuit approaches infinite attenuation.
The bridge essentially has two parallel paths from the plate output (bridge input 1) to the cathode output (bridge input 2). These signals are applied to one parallel path (say, Z1 & Z2 below); at the midpoint, the RC circuits for that side of the bridge shift the signal phase a certain amount. The same signals are applied to the other parallel path (Z3 & Z4), and at their midpoint the RC circuits on that side shift the signal phase a certain amount.
What is key is the RC components of each path through the bridge have a phase shift
which is 90 degrees apart from the other bridge path at any given frequency. In other words, take the lower graph of phase on the Filter Plot, mentally duplicate it, and slide one of the graphs to the side so that at a given frequency one output is 90-degrees away from the other.
In practice, this is accomplished by changing the cap values between the same-positioned cap in the two bridge paths. The resistances stay the same. For example, the plate output of the 1st split-load on the Vox schematic shows two paths with a cap in series with a resistor. The resistors are both 68k, but the cap values are 500pF and 2000pF (0.002uF).
The phase of the voltage across the resistor of the series circuit is given by the equation at the bottom. When I ran the numbers for 68kΩ and 500pF or 2000pF at a frequency of 1kHz, I got that the phase for the 500pF case was ~30.33 degrees, while the 2000pF case was ~77.94 degrees. Those are ~47 degrees apart.
But that doesn't account for the input from the cathode signal and the shifts found due to their cap values. You will find the result is at the brdges two midpoints, the two outputs stay 90-degrees different regardless of the frequency applied and the total amount of phase shift in the circuit.
So then the modulator's job is pretty easy: make these signals alternate between less-than-90-degrees-apart and more-than-90-degrees-apart.
If you're willing to skip to solid-state, then you can find circuits called "Quadrature Oscillators" which have 2 outputs 90-degrees apart. This usually simplifies parts count and build complexity. See pages 18-19 of this TI Application Report.