instrumentation of V-divider dual-thermistor circuit
To a first approximation, the reason the transmitter doesn't affect V1 and V2 in the same way is because the transmitter's input resistances aren't identical.

In 1, below, the left-hand side of the bridge is shown, and if  R2 is precisely 10000  ohms, E1 is precisely 3.3 volts, and E2 is precisely 1.31 volts, then RT1's (the thermistor's) resistance must be ~ 15191 ohms.

In 2, then, if E2 rises to 1.50 ohms and everything else remains the same as in 1, the parallel combination of the thermistor and the  transmitter's input resistance must be 12000 ohms.  

In 3, if the thermistor's resistance is 15191 ohms and the total resistance of the parallel pair is 12000 ohms, then the effective resistance the transmitter is shunting across the thermistor is 57127 ohms.

The math is the same and the results are similar for the the other half of the bridge, with the transmitter's input resistance on that side being 40122 ohms connected to 3.3 volts.