Removing/Subtracting Lead Resistance Using Op Amp Circuit
I'll just give an intuitive answer with some numbers.







simulate this circuit &ndash; Schematic created using CircuitLab

Figure 1. The circuit redrawn.


We'll leave out RW2 and R7 for now since there is no significant current flowing to the + input.
With 1 mA flowing through the RTD the voltages are as shown in Figure 1. Note that the bottom of the RTD is at 2 mV and the top at 102 mV. 
The + input is also at 102 mV.
The right side of RW1 is at 104 mV.
The op-amp is in an inverting mode configuration. That means that the output will move to that value that makes the - input equal to the + input (102 mV). Since the amplifier input is 104 mV and 2 mV is dropped across R5 another 2 mV will be dropped across R6 since both R5 and R6 are the same value.
The result is that the output is 100 mV which is the correct value for 1 mA through 100 &Omega;.


R7 is included to remove any errors due to op-amp bias currents. The - input bias current is feeding into two 100k resistors so a value of half of 100k is used on the + input. The missing 0.1 &Omega;  is presumably to allow for RW2.