Load cell amplification circuit operation
That is just the input stage to an instrumentation amplifier.  The voltage across \$R_g\$ is the difference in voltage between the two corners of the load cell, thus you know the current through \$R_g\$

Knowing that number, and knowing that the same current must flow throuh both \$R_F\$, you can calculate that the difference in voltages between the op amp outputs will be $$ \Delta V \left( 1 + \frac{2R_F}{R_g} \right) $$, where \$\Delta V\$ is the difference across the corners of the load cell.

The advantage is that you can change the gain by changing one resistor, \$R_g\$.  You might have to do some work to figure out what the effect of mismatched \$R_F\$ does.