How are the terminal voltages regulated for this Wheatstone bridge?
Using the bridge excitation voltage as the ADC reference gets around the problem of requiring a high precision and also low impedance excitation voltage source. The Wheatstone bridge is sensed differentially, so as long as the ADC reference tracks the excitation it doesn't matter what the exact excitation voltage is. The Kelvin connection is used so that the reference voltages are exactly the same as the excitation voltage seen at the bridge.

To expand on this, consider the circuit you've given. The differential output from the Wheatstone bridge is given by:

\$V_G = (\frac{R_2}{R_1 + R_2} - \frac{R_x}{R_x + R_3})\cdot V_S\$

To simplify this, let's assume that \$V_- = 0\ \mathrm{V}\$ and take all the resistance terms into \$A\$, which gives:

\$V_o = A\cdot V_+\$

The important part here is that the output voltage from the bridge is dependent on the excitation voltage, \$V_+\$. This implies that you need to regulate this voltage to get consistent results. It is worth noting that the impedance of the Wheatstone bridge may be quite low, so you need a high accuracy low impedance voltage source. Most accurate and precise voltage sources generally don't supply high currents, so this presents a potentially expensive design problem. 

However, the way this circuit is designed avoids this problem. Again assuming the negative input is at ground, the output from the ADC is given by:

\$D_N = \frac{V_\mathrm{IN+}}{V_\mathrm{REF+}}\cdot 2^N\$

where \$N\$ is the resolution of the ADC in number of bits. The output code depends on the reference voltage \$V_\mathrm{REF+}\$. Now, let's digitise the output from the Wheatstone bridge:

\$D_N = \frac{A\cdot V_+}{V_\mathrm{REF+}}\cdot 2^N\$

It's easy to see from this that if \$V_+ = V_\mathrm{REF+}\$ then the output code becomes independent of the excitation voltage and ADC reference voltage, so it is not necessary to provide the highest quality fixed sources:

\$D_N = A\cdot 2^N\$

This also means the output from the Wheatstone bridge does not exceed the ADC reference voltage. The Kelvin connection ensures that the excitation and ADC reference voltage are the same, so this cancellation works.