Why does a Schmitt trigger work in saturation region?
When you solve positive feedback circuits like this, you need some initial values.

We can say that \$V_{sat+}\$ as the upper limit to what the opamp can drive to and \$V_{sat-}\$ as the lower limit.

If we make an initial assumption that \$V_{out} = V_{sat+}\$ then you will get 
$$ V_+ = V_{sat+}\dfrac {R_1}{R_1+R_2} $$
$$ V_{out} = A_v(\dfrac{R_1}{R_1+R_2} V_{sat+} − V_{in})$$

When \$V_{in} &lt; \dfrac{R_1}{R_1+R_2} V_{sat+}\$ the output will be \$V_{sat+}\$
When \$V_{in} &gt; \dfrac{R_1}{R_1+R_2} V_{sat+}\$ the output will be \$V_{sat-}\$

You would do the exact same procedure with an initial assumption that \$V_{out} = V_{sat-}\$ to see when you would see a transition going in the opposite direction.

Check out page 7 of Opamp circuits - Comparitors and Positive Feedback