In practise, the efficiency of an LDO regulator is dependent on how much voltage it is dropping. While operating, an LDO is effectively a resistor that varies in real-time to ensure its output voltage stays stable despite changes in load current.

Power = I (current) x V (voltage)

Since an LDO is a resistive element, yes, it burns off energy as heat in this process. So an LDO dropping 18V to 5V could be very inefficient, more so when driving a higher current load as shown by the formula above. However, if an LDO is used to generate a 3.3V rail from a 5V rail, it is dropping just 1.7V, resulting in less power dissipation for the same load current.

You’re right, using an LDO for a large voltage drop is not good electrical design. But LDOs have excellent noise rejection performance, meaning they can take a noisy rail from a switching buck/boost converter with lots of transient or high-draw components on it, and create a much cleaner rail for lower-current, more sensitive devices. This is what I expect Apple’s doing.

LDOs drop the difference in voltage as heat, yes - hence the voltage difference between input and output determines the efficiency (eg a 3.0v LDO fed by 6.0v is 50% efficient).

In many cases where LDOs are used in designs the amount of lost power is negligible as the current being drawn is so small - simply not worth using a switcher for that rail. Also, LDOs typically have cleaner output so often an LDO is used to isolate an analog subsystem from noise on the main (digital) system rails.

+ some designs will have both a switcher and an LDO for the same power rail and switch to the LDO when the current is very low. Switchers get inefficient at low currents, so having both can improve efficiency.