LED - light emitting diode
The way it works in a nut shell, is that they have low resistance, but a voltage drop (threshold voltage) when run in the correct direction. If you plug it in the opposite way, it will not turn on.
That box in the end has a lot of stuff in side, most likely a power supply to drive the LEDs, and some resistors to keep the current in the right range. LED power draw and the brightness is based on how much current you supply it, and since it is not a resistor, you cannot just divide the voltage over its resistance and calculate how much current it is drawing.
As a back of the envelope calculation, when I was playing with LED I usually use something like a 1.2V drop per LED in series as a reference. Then I subtract the voltage from the power supply voltage (i.e. a 12V power supply to drive 5 LED will be 12V - 5 * 1.2V = 6V), and divide that difference by the resistors valve to keep it at around 20mA (reference LED has full brightness at 20mA and will last a couple decade, 10mA will reduce brightness and last longer, more than 20mA will burn it out without any extra brightness, etc), so in this example the 6V / 20mA = 300 Ohm resistor in series.
Please remember that it all depends on how many LED they wire in series, and the actual spec of the LED they use (may not be 1.2V and may not be 20mA). When in doubt, use a higher supply voltage and larger resistor in series to minimize the affect of the threshold voltage's influence in case it is off, and when in doubt, bigger resistor to limit the amount of current to below 20mA for safety. i.e. If I use 120V to drive 50 LED in series (60V) with 3k Ohm resistor will be safer than using 90V to drive 50 LED in series (60V) with 1.5k Ohm resistor. If the LED voltage drop I use was off by 50% (i.e. 30V drop only), the amount of current difference would be (120V - 30V) / 3k = 30mA (10mA over) for 120V supply vs (90V - 30V) / 1.5k = 40mA (20mA over).
Too much trouble if you just want to save space. It would be easier if you just cut the wire and extend it, and relocate the box elsewhere.