Common ground in low side current sensing
To measure current on the low side of circuit, do we still need to have a common ground between V1 and V2?

If you are absolutely sure that there are no hidden paths between the power supplies and any part of the circuit other than what is shown, you will be OK.

Do the multimeters make a common ground with the source too?

Multimeters are designed to ensure that the condition just mentioned, of no hidden paths is met. So, they often have plastic cases and battery power.
In addition, because there are no hidden paths between power supplies, it is safe for the ground (&quot;common&quot; really, because it isn't connect to &quot;ground&quot;) of the multimeter to be connected to the negative probe. This is probably done, though I certainly cannot guarantee it is done in every case. It is probably done, because it is safe (because of the &quot;isolation&quot;), and because it eliminates some unnecessary worry about accuracy and proper operation of the multimeter.

V2 can be a power source of any kind (battery or switching/linear PSU) without a connection to earth

A problem arises when the assumption that the power supplies are truly isolated is mistaken. One may naively think a power supply is isolated, though it may not be. Even a very small current can disturb your circuit, as illustrated by the circuit below, in which 120AC &quot;leaks&quot; through a pair of capacitors between one rail of V1 and ground.



simulate this circuit &ndash; Schematic created using CircuitLab
The output, and the node marked &quot;cm&quot; are shown here. The &quot;cm&quot; node is the common mode signal present at the inputs of the op amp (assuming the op amp is still operating in its linear region).

One can see that the output appears to be a solid 2V. However, the peaks of V_CM are at about 10V. If the capacitors were just a little bigger, the peaks of V_CM would reach the rail voltage of 12V. A little higher and the op amp will work incorrectly, giving unknown results at the output -- probably one of the rail voltages. (The simulation does not show this, happily treating voltages at the inputs which are above the rails as if nothing untoward was happening.)
How much current was required by the noise source to bring this circuit to its limit? Only 1mA peak. That may seem very insignificant compared to the 10A that is flowing through \$R_{sense}\$, but it clearly is not insignificant.
The details of when a common mode noise source will drive the op amp inputs beyond the rail voltages depend greatly on the voltage of the common mode noise source, it's impedance relative to the circuit under test, and the resistors in the diagram. Change any one of those, and one will get different results. Different parameters, and any conclusions about immunity to noise need to be re-evaluated.
The impedance of 2 30nF capacitors in series at 60Hz is about 177k\$\Omega\$. Again, the current created by the noise source is only 1 mA peak, but the common mode signal at the op amp is 10V peak.
That is what could happen, if your voltage source is not truly isolated. However, if, like a portable ammeter, the voltage source is a battery, and the case is plastic, you can be quite confident that the impedance seen by your circuit by any common mode noise source will be much higher, (or the voltage of the noise source much lower) and your circuit will measure the voltage across \$R_{sense}\$ just fine.