LEDs (all diodes) are non-linear devices. If you look at a voltage-current curve, you will note that the relationship is not a straight line as it is with a resistor:
From 0V, as voltage increases, not much happens until the forward voltage (Vf) of the diode is reached. After that point, current increases exponentially. If there is too much current, the LED will fail.
It is for this reason that current through an LED needs to be limited. (For example, with current-limiting resistors.)
The Vf point for LEDs differs. Even for two identical LEDs, this point may be slightly different due to manufacturing tolerances. The one with a slightly lower Vf will inevitably have higher current. In the long term, the one with higher current will fail before the other, and start a cascade failure.
simulate this circuit – Schematic created using CircuitLab
In the above schematic, imagine that the LEDs have a Vf of 2V. The desired current through them is 20mA. If they drop 2V, then that leaves 3V to be dropped across a resistor. Current through the LEDs in parallel will be combined in the resistor, thus 75Ω is calculated to limit current to 40mA.
In this scenario, it could be the case that D2 has a slightly lower Vf, and actually receives more than its 20mA maximum. While this might not cause any apparent damage in the short term, sustained current exceeding its limit will eventually cause it to fail. When that happens, D1 will "see" all 40mA, and fail much more quickly.
simulate this circuit
This is a better way to limit current to two LEDs. If D2 fails for whatever reason, R1 will still limit current for D1 appropriately.
This effect should be even more apparent when you have two different diodes, where the Vf is much less similar between them. In parallel, the diode with the lowest Vf will conduct more readily and current through it will increase quickly.
Imagine that, in your schematic, D1 Vf is 1.8V and D2 Vf is 2.1V. (Q1 on-resistance won't be 0, but for simplicity we'll treat it as such.) With 5V applied, D1 will drop 1.8V. R2 will have 3.2V across it, and limit current to 14.5 mA. Assuming that 14.5 mA is no problem for D1, it will light. However the voltage "seen" by D2 is only 1.8, which isn't enough to cause it to light.
If you increase voltage, you effectively only increase the current through D1, why? Because it still has a Vf of 1.8. Imagine that you increased the supply voltage to 10 volts. R2 will now have 8.2V across it, limiting current to 37mA. If D1's maximum current is 20mA, it will very soon fail. Once that happens, voltage will rise to the 2.1V necessary to light D2. Now the voltage across R2 will be 7.9V and current will slightly change to 36mA. If D2's maximum current is 20mA, well, you can see that it fails, too.
This is an oversimplification, but hopefully it will demonstrate why LEDs in parallel behave the way they do.