How to calculate the amount of change on gate-source voltage needed in this two stage comparator?
In the current conditions, \$I_7\$ is not equal to \$I_6\$. Let's write the equation for \$I_6\$:

\$I_6=\dfrac{1}{2}*K_6*\dfrac{W_6}{L_6}*(V_{GS6}-V_T)^2\$

\$V_{GS6}=\sqrt{I_6*\dfrac{2}{K_6}*\dfrac{L_6}{W_6}}+V_T\$

When we change \$V_{GS6}\$, \$I_7\$ will be equal to \$I_6\$.

\$I_7=I_6=\dfrac{1}{2}*K_6*\dfrac{W_6}{L_6}*(V_{GS6}'-V_T)^2\$

\$V_{GS6}'=\sqrt{I_7*\dfrac{2}{K_6}*\dfrac{L_6}{W_6}}+V_T\$

\$\Delta V_{GS6}\$ is the difference of the two conditions and can be written as:

\$\Delta V_{GS6} = |V_{GS6} - V_{GS6}'|\$

\$\Delta V_{GS6} = \sqrt{I_6*\dfrac{2}{K_6}*\dfrac{L_6}{W_6}}+V_T - \left(  \sqrt{I_7*\dfrac{2}{K_6}*\dfrac{L_6}{W_6}}+V_T\right )\$

\$\Delta V_{GS6} = \sqrt{\dfrac{2}{K_6}*\dfrac{L_6}{W_6}}*\left(\sqrt{I_6}-\sqrt{I_7}\right)\$

This result is the same thing as the equation you have given:

\$\Delta V_{GS6}=\dfrac{\sqrt{I_6} - \sqrt{I_7}}{\sqrt {\dfrac{K_6'}{2}*\dfrac{W_6}{L_6}}}\$