There's nothing strange going on here.
With a 3.3V supply, and a blue LED, there is very little voltage across the 220 ohm resistor(s).
simulate this circuit – Schematic created using CircuitLab
Your Fluke meter (when set to measure current) has quite significant internal resistance, which is higher on lower current ranges. This resistance adds to the 220 ohms/n (where n is the number of LEDs), and explains the lower reading on the lower current ranges. There are times when you can ignore the voltage drop of an ammeter (or the loading of a voltmeter) but this is not one of them.
Another effect is that the output port pin also has some internal resistance (which again adds to the parallel resistors) so you expect the current to drop if you're driving the LEDs with a single port pin.
I suggest you get a better current rated transistor, or even a good old MOSFET (much better, you don't have enough voltage headroom to use the transistor properly anyway.. who knows how the sparkfun guy got stuff to work at all). For what you want to do, get a 2-3A rated MOSFET, or at the very least a 1.5A+ rated NPN transistor.
It is not a good idea to operate 3 LEDs at 1.6-1.8V on 5V, and expect a 2Ohm resistor to regulate current properly. The variation in forward voltage is too much, and having such a small resistance (also with poor tolerance) you will not get very good results.
I suggest you use 2 LEDs in series on each chain, and use a larger resistor. To get 100mA out of 1.4V spare (3.6V out of 5V is taken up by 2 LED in series) you need about 14 Ohms, which is surely better than 2 in terms of leeway for tolerance. The other thing is, both 2 and 14 ohms are unusual/non standard values, you might need to find a nearest standard value. Also remember your LEDs should only be on for the picture, for a short period of time, so it's not actually that bad if your LEDs run slightly over-current.
The LEDs used by the Sparkfun tutorial are 1.6-1.8V x 10mA, meaning they are only really 18mW each, and there are 13 LEDs. That is 13 x 18mW = 234mW total of IR light. You are trying to do 25 LEDs at 1.6-1.8V x 100mA, meaning your IR light output will be a ridiculous 4.5 Watts. Do you really want x20 more IR light than the tutorial guy had? I don't think you really thought about any of this..
The basics for calculating R6 in your case is if you do end up using a NPN transistor, the base current into the transistor is determines how much current flows through it. Your LEDs do the current limiting, so there is no real reason to even use a transistor (which effectively act as current-amplifying switches). The correct component for this digital on/off functionality is an N Channel MOSFET. Both should have a base/gate resistor though, but the MOSFET one is almost not needed, rather it's recommended. It can be something simple like 100 Ohms.
The base resistor for a transistor allows you to control current through the collector-emitter using the DC current gain/beta factor of the transistor, usually shown on the datasheet. If you have a gain of 100, it means 1ma into the base will allow 100mA through the emitter-collector. Problem is, as the base current approaches saturation, the current gain drops dramatically until it is quite low, like 10 or so. This is different for each transistor however. If you put 40mA into the base, it will probably saturate, causing the transistor to act more like a switch with minimal forward voltage drop, which is what you would want to happen in this LED driving application.
UPDATE: based on the feedback from OP, I have provided the below diagram to show the correct way to hook up the IR LEDs, with a low-side NFET power switch. Note the FET should have a "logic level" gate drive voltage, around 2V threshold should be good for 3.3V control. The FET should also be rated for 3+ Amps. I believe it was worked out to be about 800-900mA continuous, for 4-5 hours in this user-case scenario.
simulate this circuit – Schematic created using CircuitLab
Best Answer
To answer your first question, yes. 3.3V is not enough to forward-bias two LEDs, which typically need somewhere around 2V each for red LEDs and more for other colors.
For the second question, putting them in series is entirely reasonable, and often preferred as it prevents current crowding (where one LED with a slightly lower forward voltage "hogs all the current").
As for the third question: You'd have to measure the forward voltage of the LED. Many, if not most, multimeters can measure the forward voltage of a diode, but few (at least that I've seen) can measure voltages as high as most LEDs. Not using a resistor will damage your LED, your rpi, or both, and is not recommended.
To measure the LED forward voltage, try the following circuit:
simulate this circuit – Schematic created using CircuitLab
This will give you an idea of the LED forward voltage. If you hook up this circuit and the LED doesn't visibly turn on (which may happen for high-current blue or white LEDs), try changing to a 100Ω resistor.
A better circuit would involve an active current source, but this one is probably good enough for your purposes.