The TMC drivers limit the current to the value you set, which is most of the time lower than the datasheet. In no case you will need more than 1.3 A * sqrt(2) * 2 motors = 3.6 A. In fact, you won't need this much either.
When there is no field in a coil, the driver applies full voltage, but the current is low (initially zero) so you don't hit the 1.3 A per coil.
The current increases (the higher the maximum voltage available to the driver, the faster it increases) and the driver (probably) keeps the full voltage until the preset current is reached. Just a moment before that, the current is almost there, but you still have full voltage from the power supply. This is the theoretical worst case, but it applies only for a very short amount of time.
As soon as the current reaches the preset, the driver "cuts the voltage" to keep 2.4 ohm * 1.3 A = 3.2 V (because V = R * I). This means that the power supply sees 3.2 A/12 V*1.3 A = 0.35 A.
When running, the motors almost never start from zero to max current: both coils are powered and when one increases, the other one decreases.
In fact, the microstepping makes the steppers act more or less like AC motors with two phases. This means that overall the current is the max current per phase multipled by sqrt(2). Also, when using microstepping one phase (coil) is not completely shut off, but two of them work at the same time (with different current levels). This means that in total one compensates the other, and the power supply only provides, more or less, 0.35 * 1.4 = 0.5 A per stepper. You have two, so it's 1 A total, therefore 2 A PSU considering the inefficiencies.
A very easy and complete explanation is provided here:
By controlling the duty cycle of the chopper, an average voltage and
an average current equal to the nominal motor voltage and current are
As the current increases, a voltage develops across the
sensing resistor, which is fed back to the comparator. At the
predetermined level, defined by the voltage at the reference input,
the comparator resets the flipflop, which turns off the output
transistor. The current decreases until the clock oscillator triggers
the flip-flops, which turns on the output transistor again, and the
cycle is repeated
So you never have coming out of the PSU more than the preset current.
Supply current is not the same as the motor current in a copper drive. It is the motor current multiplied by the dutycycle, at
standstill typically Isupply = IM · ( VM ⁄ Vsupply )
how the H-bridge is switched during the turn-off period, the current
will either recirculate through one transistor and one diode (path 2),
giving the slow current decay, or recirculate back through the power
supply (path 3). The advantage of feeding the power back to the
power supply is the fast current decay and the ability to quickly
reduce to a lower current level.