This is a good question, many people in the business do not understand the difference despite its importance. Unlike a brushed DC motor, the current drawn from the battery can be quite a different value than the current measured at the motor phases.

Battery current is always a direct current (DC) value. This means that you can take the battery current value, multiply it by the battery voltage and arrive at a battery power value. Example 48V X 50A = 2400W. This is easy and handy which is why it is used on the sales side of things.

Brushless DC motors are actually three-phase synchronous (no slip between rotor and rotating stator field) AC motors, the only reason they are labelled DC is that they are powered by a battery. The controller (sometimes called an inverter) takes the DC power and used high speed switching devises such as MOSFETs to generates three phases at current levels that can be controlled (to modulate torque) and at a frequency that can be controlled (to modulate speed). The motor phase current is very closely related to the motor torque. Calculating the power at the controller output is more complicated than calculating the power at the battery output but it is still the product of the phase voltages multiplied by the phase currents. What makes things more complicated is that the current waveform is a sinusoid and the voltage waveform is something else (hopefully something kind of close to a sinusoid) they may be out of phase. Usually when we talk about motor current, we are talking about the peak value not the RMS value. If the waveform is a perfect sinusoid the RMS value = 0.707 X peak value.

In instances where high torque is demanded of the motor but the speed is low, the phase current can be very high (torque is proportional to motor phase current) but the power can be low because the motor phase voltage (back EMF) is low. Since the law of conservation of energy applies, the power drawn from the battery is roughly equal (the controller is about 97% efficient) to the power sent from the controller to the motor, the low power results in a battery current draw which is much smaller than the motor phase current because the battery voltage is always about the same.

On the following page I have shown some waveforms from a simulation of the ASI IPM motor at 291A (peak) motor current and 500 RPM. Since the speed is low the motor phase voltage is only 6V (peak).

Determining the motor electrical power
involves multiplying the instantaneous motor current by the instantaneous motor
voltage in each of the three phases and integrating that over one commutation
cycle (the blue line in the last graph).

If the controller in this instance is powered by a 48V battery the battery current will be 2100W/48V=44A DC

Notice the 44ADC is occurring at the same time that the motor current is 291A (peak). There is no simple formula to link the two.