Why does my current in parallel circuit not add up when I check using a multimeter?
Let me try to explain it without much math and physics. For the sake of the principle, let us assume that the multimeter uses 25Ω shunt resistance (not unreasonable for a 40mA range), everything is perfect and the LEDs are dropping 2.7V. Then we can use the following LTspice simulation to illustrate what is happening:



With setting the shunt resistors to 1nΩ in simulation (basically off) we can determine a perfectly equal 1.0487mA per LED (it is a perfect simulation after all), thus a total of 2.097mA.

Let us now see what happens when we put the Shunt1 to 25Ω: The voltage drop across it reads 52mV which corresponds to 2.08mA. Already not quite what the simulated perfect value is. But wait, there is more. Let us set Shunt2 to 25Ω resistance and measure the voltage drop over it. It will read 22.98mV or .919mA only!

Why is that so? Well, since we have a simulation here, we can poke around and look at the "real" values, and see that through the other diode in that configuration, there is a total of 1.175mA flowing. There is our "missing" current. In the measurement situation you have created a different circuit than the one you were trying to characterize, thus you can not compare those too. Do the calculation for the above circuit where Shunt1 is 0 and Shunt2 is 25Ω, and you will see that the voltages and currents are distributed differently. 

Now what you can do is to either buy some measurement equipment with much lower burden voltage, or you can figure out the shunt resistor size used in your multimeter, place it at the position of Shunt3 when you measure Shunt2, and with only minimal impact on the overall current flow (assuming 25Ω is in the right ballpark) you can measure things more properly now since the current won't get "redirected".

Btw. this is also the reason why you measure different values for different current ranges: your meter simply uses differently sized shunt resistors there.