How to determine terminal characteristics of a dc motor with known torque and angular velocity

In the steady-state case, it's a simple linear interpolation:

\$I_S\$ is the stall current.

\$\omega_{nl}\$ is the no-load speed.

\$I_{nl}\$ is the no-load current.

For a motor loaded such that it spins with a speed \$\omega_L\$, the current draw is:

$$I_L = (I_S- I_{nl}) (1-\frac{\omega_L}{\omega_{nl}}) + I_{nl} $$

For example, you have a motor with a 2.2A stall current, 0.2A no-load current, and let's assume the no-load speed is 1000 rpm. Technically \$\omega\$ usually refers to rads/s, but here the units cancel out; the only important factor is what the relative speed of the motor compared to the no-load speed is.

At a 250 rpm load speed, the current draw is:

$$I_L = (2.2A - 0.2A) (1-\frac{250}{1000}) + 0.2A = 1.7A$$

Keep in mind that even though the steady-state current of the motor is 1.7A, the peak current could be much more. Starting up the motor would draw close to the stall current of 2.2A, decreasing until the motor reaches steady state. Depending on how quickly the motor is able to get to steady state, this may be significant or not.

Looking at the datasheet, while it doesn't lie, it is definitely on the edge of misleading.

Note that the "max power" stated occurs at almost exactly half the unloaded RPM and half the stall current. This is indeed the "max power" point for such a motor, but the datasheet fails to mention that it is also the nominal 50% efficiency point, thus dissipating 12V*68A-337W = 479W in that tiny motor - destroying it, probably in minutes.

(Ideally, exactly half the power would be delivered, about 400W shaft and 400W heat, but the motor isn't ideal).

The motor is probably suitable for 100-150W continuous output and 200-250W short term.

So practically you must operate the motor at the upper end of the speed range, and if the speed falls below (say) 70% of the unloaded speed (or the current rises to 30% of the stall current) then - unless this is strictly temporary, like starting a heavy load or hitting a chilled spot while machining a cast iron surface, you need to cut the current and protect the motor.

Then the question of which side of the torque speed curve doesn't apply - unless the protection has tripped, you should be on the high speed side.

You can get circuit breakers that will allow short-term overcurrent. These are "motor rated" or Class C breakers for the AC motors used in most machine tools. I don't know of anything suitable for 12V DC though. I'd be looking for a 12V DC supply that can be set to trip if its output exceeds 40A for more than a couple of seconds. And as Olin says, if you want to monitor it yourself, measuring the current is definitely the way to go.