Why does POT cause current to divert?
Almost certainly, the two LEDs are not perfectly matched. For the same current (and brightness) level, LED1 has a greater voltage drop than LED2. Additionally, you are using confusing terms - what you call "open" on the pot is actually what everybody else calls "shorted", that is the resistance is almost zero. Just for illustration purposes, let's say that LED1 will operate well at 2 volts, while LED2 will shine at 1.5 volts.

So, at first most of the current is flowing through LED1, and it produces 2 volts. At the same time, LED 2 is being driven by LED1's 2 volts through the pot, to the nominal 1.5 volts. The pot is in the high-resistance position, so not a lot of current gets to LED 2. As a result, LED1 is bright and LED2 is dim.

Now turn the pot to what you call "open", which is the low-resistance setting. Now LED2 is hogging most of the current, since it operates at 1.5 volts, and this is not really enough to light up LED1. The result is that LED2 is bright and LED1 is dim.

If the two LEDs were identical, when the pot was set to "open", the two LEDs would be of equal brightness. But they're not.