If you really want two thresholds, you won't be able to tell which one tripped unless you get rid of the diode OR and use two Schmitt triggers.
In the general case, you need to select R13 and R14 so that the voltage at their junction does not exceed the supply voltage (VCC) on the Schmitt trigger. If you make them equal (you suggested 10k), then you need a VCC of at least 15v (+VM/2). Another option is to connect a diode from their junction to VCC, cathode to VCC, to limit the voltage swing. Having said all that...
[edited to distinguish between Vtr and Vtf]
For the trip point Toc (overcurrent sense),
Rp*C6=-Toc/[ln(1-Vtr/Vf)]
Where Rp=R13*R14/(R13+R14),
Vtr is the rising Schmitt threshold voltage,
Vf=Vm*R14/(R13+R14).
This assumes no OR'ing diodes, and no R17.
Select the resistors first, calculate Rp and Vf, then calculate C6 to give you the desired Toc.
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For the release point Trel (overcurrent condition has ended),
R14*C6=-Trel/[ln(Vtf/Vf)]
where Vtf is the falling Schmitt threshold voltage.
Or, solving for Trel,
Trel=-R14*C6*ln(Vtf/Vf)
Because R13 is not involved in the discharge time constant.