Some motor basics.
A motor's stall current is determined by the resistance of the coil. (quoted at the nominal operating voltage). If the shaft is clamped so it cannot rotate then the 'motor' will act like any other resistor and follow Ohm's law - so yes, the stall current will increase with increasing applied voltage.
Once the rotor starts to rotate it will induce a back emf so the voltage 'seen' across the rotor will be reduced. The net current through the motor will be reduced.
Increasing the applied voltage will increase the speed (how much depends on the motor). With increasing speed comes increasing loss (friction, windage etc.) so there will be a corresponding increase in current.
Applying a load (torque) to the output shaft slows the rotation speed and reduces the back emf. This in turn will increase the motor current. Applying too much torque may reduce the shaft speed to 0 and this is once again back to a stall.
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.
Best Answer
It's because, for a DC motor, the torque curve and speed curve are both linear and of opposite polarity slopes and both curves also start and end at the same place. 100% torque = 0% speed, and 100% speed = 0% torque.
Those two facts mean that the adding them up will always equal 100%. Specifically, 50% of one corresponds to 50% of the other, and since power = torque x RPM, this just happens to be when the multiplication of the two is the maximum.
At one extreme end, you have 100% torque but 0% speed so the multplication gives you zero output power.
At the other extreme end you have 100% speed but 0% torque so the multplication again gives you zero output power.
Geometrically, it is identical to why a square is the rectangle with the most area if the sum of all sides is a fixed number. If one side is speed and one side is torque with power being the area, and the sum of length and width must be 100, a square of 50 x 50 gives the most area.