Understanding why not to use a resistor for multiple LEDs
LEDs’ forward voltage drop varies from device to device and by more than you might think. This causes a few issues for the user.

Figure 1. A Lite-On LTST-C170TBKT. Image source: Lite-On.
The Lite-On LTST-C170TBKT InGaN blue LED datasheet, for example, shows the forward voltage varies between 2.8 V and 3.8 V at 20 mA. This is quite a variation. The lower value would, just about, allow operation with a low value series resistor at 3.3 V whereas the 3.8 V sample might not turn on very brightly. On higher voltages control of the current by series resistor would vary for different LEDs.

Figure 2. LTST-C170TBKT-Vf-variation is rather wide. Image source: Variations in Vf and binning.
Here notice that if we applied 3 V to a group of these LEDs some would draw one milliampere and others could draw 50 mA.

I don't understand this: On an LED Voltage-Current graph, does the LED &quot;pull&quot; a certain current like it has resistance?

Resistors don't &quot;pull&quot; current. They allow a certain amount to pass and the amount (I) is directly proportional to the voltage (V).

Figure 3. You can consider any diode as being a little like a non-return valve with a spring. Image source: What is an LED?.
You may be able to imagine that if you paralleled a set of many of these non-return valves with assorted spring pressures that as you raise the water pressure the one with the lowest resistance would open first and hog all the current. The ones with higher spring pressure would only open a crack. The first to open will determine the pressure drop across the whole lot. In the same way the diode with the lowest forward voltage, Vf will conduct first and it will determine the voltage applied across the parallel group.

If I have 2 LEDs in a parallel circuit, both parallels have to drop 2V, but one LED is 2V at 5mA, and the other is 2V at 10mA, this does NOT make sense to me.

Hopefully it does now.

I have a resistor before the circuit, lets say from an Arduino 5V, that drops 3V (to give 2V to the LEDs) at 15mA, so 200 ohms [small 'o']. This is what I don't understand: Wouldn't the current be shared between the two, giving 7.5mA to each?

No. Due to the differences in Vf one will conduct more than the other.

What is the usage of the 'voltage versus current' graph? My guess is to show what equal lighting takes.

It shows the relationship between current and voltage for a typical LED of that type. Figure 2 may be more useful.
Have a read of my linked articles. They may help.

From the comments:

&quot;others could draw 50 mA&quot;, I thought it was the resistor, not the LED, that allowed the current through, though?

In my 3 V LED discussion I am applying 3 V directly to the LEDs, not through a resistor.

Also, having the different forward voltages doesn't match my understanding of how each parallel branch drops the same voltage.

As Figure 2's blue line nominal curve shows the Vf is not a constant. Once it gets into the linear region it is slightly off vertical so small changes in the applied voltage result in large changes in current. Now imagine a bunch of blue lines all slightly offset to the left and right of the one drawn and you should be able to see that a vertical line drawn from any point on the V-axis (representing the common voltage applied to the parallel LEDs) will result in different currents in each of the parallel LEDs.

Does the current through one branch increase because the wires have to drop the voltage?

Generally not. It's because each LED's curve will be slightly different. There will be tiny differences due to the internal connecting wires but these are metallic, not semi-conductors.