meaning of floating supply voltage
The floating point is the Vs terminal. The power supply for the high-side driver is the bootsrap capacitor, installed between Vb and Vs. This voltage is floating, that is its reference point, Vs moves with respect to COM. That happens when the high-side MOSFET you drive turns on; its drain is connected to a positive supply rail and when it turns on its source goes to that supply rail, so Vs has to follow the MOSFET source and Vb will be higher than Vs by (Vb-Vs) volts.
There are two undervoltage circuits. The upper UV DETECT block is referred to Vs, the other to COM. The reason is that you drive one MOSFET with respect to Vs, the other with respect to COM, so you need to make sure that the drive voltages with respect to THESE points are correct, since the sources of the MOSFETS are connected to these points and the gates have to be more positive than these points.
Now the upper driver can be considered separately powered by the boost capacitor, connected between Vb and Vs. Then Vs can be any potential with respect to COM. The trouble is, your drive signal is resferred to COM, so how do you tell this driver to turn on the MOSFET? They use two resistors, connected to Vb, as drain resistors for two MOSFETs, having their sources connected to COM. Since the sources are at COM potential, they can be driven by a signal referred to COM.
The drain curents of these MOSFETs will develop voltages across those resistors. Let's assume, for the sake of the argument that the voltages developed across the two resistors are equal to (Vb-Vs) volts. Then the bottom of those resistors will swing between Vb and Vs (with respect to Vs). As far as the upper driver is concerned, that is all it needs to work, two voltages varying from Vs to Vb. For this driver, Vs is the "ground" and its drive voltages are with respect to this "ground".
Try simulating or even breadboarding a simple circuit with a bipolar transistor. Measure the voltage across R2, it will be always constant, but with respect to Vb, no matter how you adjust V2 (within limits; this circuit is just to illustrate the point). In the original circuit, since Vb-Vs is constant (you have a capacitor between those pins), the voltage can also be said to be referred to Vs.
Think of the capacitor between Vb and Vs as a battery.
The circuit powered between Vb and Vs contains the flip-flop and the actual driver, its own UV block, etc., but now Vs can be at hundreds of volts above COM. The R-S flip flop is needed because the MOSFETS would dissipate too much if they conducted current all the time, so they just pulse the currents and use the flip-flop to "memorize" the last command and turn on or off the external MOSFET.
In the circuit shown, if you connect/ disconnect V1 you will also generate pulses across R2. Try it and see what happens as you adjust V2.
That's about it.