The simple reason is that LEDs are not light bulbs. An LED is not a passive device and does not follow Ohms law.
To a first approximation if the voltage across an LED is below the forward voltage then no current will flow. In that situation the voltage across the LED pins is whatever the supply voltage is.
If the voltage across an LED is over the forward voltage then an unlimited current can flow and the voltage between the pins is fixed at the forward voltage.
This approximation isn't perfect but 90% of the time it's good enough.
So with 1 LED and no resistor you get 3V over the LED and then 2V left over with zero resistance, I=V/R = 2/0 = infinite current flow. The LED is dissipating 3*infinity watts of energy and instantly vaporizes. Clearly this isn't exactly what happens in the real world but the result is enough to tell you that bad things will happen.
With 1 LED and a 200 ohm resistor you get 3V over the LED and then 2V extra with 200 ohms resistance, I=V/R = 2/200 = 10mA. The LED is within its operating current and will light up.
With 2 LEDs and no resistor you get 3V on one LED and 2 V on the second (or 2.5 on both if you prefer). Either one or both LEDs are below their forward voltage and so no current flows. The LEDs remain off.