How exactly LCDOUT statement works?

[HenrikOlsson]

Again, $FE is the "signal" to the compiler that the next byte should be sent to the LCD as a command and not as data.

LCDOUT $FE, 0, 1 will send the value 0 to the instruction register and then the value 1 to the data register of the LCD controller.
LCDOUT $FE, 0, $FE, 1 will send 0 and 1, both to the instruction register.

On the LCD controller the RS-pin is what controls which register the data ends up with. $FE tells the compiler to generate code that pulls the RS-line LOW for the duration of the next byte transfer. It's as simple as that and it feels to me like your reading too much into it.