What can the reason be to choose this specific op-amp as a Comparator/Schmitt trigger

Short:

• Yes, it's a Schmitt trigger.

• Yes, you have made a mistake.

• Output low / high happens at Vin =~ -10V*

• Output high / low happens at Vin =~ -15V*.

*I have slightly adjusted Vout range to make sums easier. Result is essentially the same. As shown the zener is being used in an unapproved manner.

Longer: Loosely a Schmit trigger can be considered to be a comparator with hysteresis or positive feedback such that the switching decision point alters when the circuit ir triggered to provide a "deadband". On that basis this IS a Schmitt trigger circuit.

Using a zener to clamp Vout as shown is a VERY bad idea.
The opamp is using all its output to try to drive the zener high. This is a non-approved way to do what is desired BUT OK for working through the example

Without using your notes.
Vout swings from about -0.8V to about 4.3V. For practical purposes we can work with 0V and 5V and get an OK answer to the main question.

Vref on the inverting input (OA-)is -5V. This does not change.

Vout is low = 0V when the non inverting input OA+ is below -5V and high = +5V when the non inverting input is above -5V.

When Vout = 0V then the trigger condition occurs when Vin = -10V, so OA+ = -5V.

When OA+ rises above -10V Vout rises to +5V.
So immediately OA+ rises to -2.5V (half way between Vin at -10V and Vout at +5V.
So the opamp is now ell away from the switching point.

To switch back OA+ needs to fall to -5V and to achieve this Vin will need to fall to -15V (as OA- is now +5 - (5 - - 15V) = 5 - 10 = -5V.)

SO the switching boundaries are
-10V rising = output low to high and -15V falling = output = high to low.

This might not be exactly right, but because the comparator has an open-drain output, you have to think of it as being something resembling the device surrounded by the dashed box:

^{simulate this circuit – Schematic created using CircuitLab}

When M1 is off (high output), the voltage at OUT is determined by the R1/R2 voltage divider. An open drain output does not drive a high voltage. Instead, it shuts off to a high impedance state, and the high voltage is created by a pull-up resistor (either supplied by you, or perhaps one that is provided in internally ("weak pullup" type feature)).

If you look at the hysteresis example in the datasheet, it works differently. The output is bootstrapped from the input. If we draw it with our explicit open drain model, it looks like this:

^{simulate this circuit}

So here, if OA1 cuts off the output transistor, then OUT is at the same voltage as IN. It is pulled up to IN by the two resistors. And of course if the transistor is fully on, then OUT is pulled to ground.