If you look at the diagram, 3,5,8 all go only to the bandgap reference and is marked "internal function". I don't see what else this could be than a Kelvin type connection to the low side of the bandgap reference. It's clear to me that the 2 mA can not go through 3,5,8. These 2 mA flow because of the resistors in the feedback loop, and come out through ground pin 4.
Let's analyze the circuit. We have a 1.25 V bandgap reference and a Kelvin current shunt of 1.25 ohm. When there's a 1.25 V voltage present over the shunt, we know 1 A flows through the resistor. This is the obvious basic operation of the circuit. But the problem then becomes, where are these things actually referenced from? I would like to refine the function of the circuit to "the goal is to keep the voltage over the shunt sense pin identical to the voltage over the bandgap reference sense pins".
This is where derive that 3,5,8 instead of all the chip's ground should be connected to the shunt sense pin. There will be a small voltage drop over the shunt's current and sense pins, and the bandgap reference's lower connection will now (presumably) be lifted from the circuit ground by that many mV. If we didn't do this, we would have an error current. This error current is not caused by the shunt's voltage drop, but by the quiescent current and the drive current for the force pin, through the metal between the chip and the lower shunt sense pin. I'm even willing to bet that this may have something to do with the oscillation.
And likewise, keeping the other two legs of the circuit identical is done by the opamp in the voltage reference chip. So what does it use as it reference? Depends on your definitions. It's ultimately using the device ground as its reference. However, if the shunt's sense pin raises the reference ground to say 0.001 V over the circuit ground, then the bandgap reference will now output 1.251 V over circuit ground which is now the amplifier's target for the shunt's higher pin (which it tries to control by driving the MOSFET.) But the goal to keep the voltage over the sense pins of the hunt at 1.25 V is still met.
So yes, I think this is the way to go. Can we get an empirical confirmation from some crazy bloke down under?