Create a differential output signal
Gain: 


I believe the gain in this design is too low to get a +/-10V output signal. 

$$V_{out diff} = \frac{390Ω}{390Ω}.(V_{i+} -V_{i-})$$
$$V_{out diff} = 1.(5 - 2.5) = 2.5V_{diff-mode}$$
or 
$$+/-1.25V_{comm-mode}$$

Alternatively you could make use of the Vocm, instead of the V- input to offset the ADC value.


Question 1:


Correct, this would destroy the DC component and is only possible if you could guarantee that the ADC generates a continuous AC signal, which I presume is impossible in your application. Even if that would be the case, you still have to take the lower cut-off frequency into account. Signals below 1Hz would require considerable capacitor values. 


Question 2:


You would measure an in-phase signal with half the amplitude of the differential mode signal on the +Vout pin and an opposite-phase signal on -Vout. 

You may want to read this application note as well. 

PS: You may want to consider a fixed voltage divider for R25/R26 in series with a smaller value potentiometer if calibration is needed.Like this for instance.



This allows for more accurate calibration. The current single potentiometer design is very course.