Some of your specs are contradictory, but I take it you want to run 2A thru a IR LED for short periods of time controlled by a 5V digital logic signal, and you have 12V car power available for this circuit. I'm also assuming this thing doesn't need to be super accurate.
Here is a simple solution, with some cautions:
This uses the inherent property of bipolar transistors where the collector current is largely independent of the collector voltage. When the base of Q1 is raised to a particular voltage, the emitter will follow about 700 mV less. This puts a fixed voltage on R1, which therefore causes a fixed current to flow thru it per Ohm's law. Most of that current will flow thru the collector, so Q1 then acts as a voltage controlled current sink. The TIP41 is a medium power transistor and has a minimum current gain of 15 in this case. That means 1/16 of the R1 current comes from the base and 15/16 of it from the collector.
You want about 2A to flow thru the LED when on, so that means the base current needs to be up to 135 mA. That's too much to expect a normal digital output to supply, so that's why Q2 is there. It provides a additional current gain of at least 50, so now the digital signal only has to source 3 mA or so, which most digital outputs can be expected to handle. Q2 and Q1 together form what is called a "darlington pair". Together they act like a high gain transistor, but with twice the B-E drop of a single one.
When 5V is applied to the base of Q2, about 4.3V will be on the base of Q1, and about 3.6V on R1. This is why R1 is 1.8 Ω, because it will draw 2A at 3.6V.
Now for the cautions. This circuit is only intended for short pulses with low average duty cycle. Significant power will be dissipated when the LED is on. With a 12V source and considering the 2A that will flow, 24W have to go somewhere. A little (3.6W or so) goes to driving the LED as intended, about twice that (7.2W or so) will heat R1, and most of the rest (13W or so) will heat up Q1. This is not a sustainable condition, particularly for Q1.
All these parts should be able to handle that power for short bursts at a time, but the average must be significantly less. If the bursts are short and the duty cycle low (10% or less), then probably nothing further needs to be done.
If you need to tolerate more power, then the first thing to do is to relieve the stress on Q1 since that's where most of the power is going. Adding a heat sink to Q1 is one approach, and reducing the voltage accross it when the 2A is flowing is another. At 12V total, 3.6V accross R1, and 1.8V accross D1, that leaves 6.6V left for Q1 to drop. Q1 in this darlington configuration needs at least 1V, better leave 1.5V, for it to do its job. That means there is about 5V extra that could be dropped by a resistor in series with the LED without effecting the operation of the circuit. That means you could add up to 2.5Ω in series with the LED to take a good fraction of the heat that Q1 would otherwise have to dissipate. The same power will still be wasted and converted to heat, but generally this is easier and cheaper to deal with in resistors than in active parts like the Q1 transistor.
Ideally the LEDs should be driven by a constant-current circuit. This will maintain a constant brightness and color as the battery drains, or as the LEDs heat up or cool down.
But the real world isn't ideal, so you can often get away by just using resistors. Yes, you should definitely use them. The resistors are there to limit the current through the LED and keep them from overheating and burning up. A 9V battery has a fairly high internal resistance, so you may be able to get away with two in series and no resistor, but it will be unreliable (changing to a different brand of battery could be enough to blow out the LEDs, etc.)
For the worst case of two white LEDs in series running at 20mA, the lowest forward voltage shown in your link is 3.2V, so you would have (9 - 6.4)/.02 = 130 ohms. The current is low, so a 1/4 watt resistor will be fine. Select the closest value to this you can find. Running at 20mA the LEDs will be pretty bright and this is a benefit: as the battery drains or the LED forward voltage changes, the apparent brightness probably won't change that much. Human vision is more sensitive to dim lights and it's harder to tell that a bright light has changed 10% than a dim light has changed 10%.
Best Answer
If you have trouble understanding what terms like 'current', 'series', and 'parallel' mean -- or have trouble understanding these diagrams -- I recommend trying out Khan Academy's excellent electrical engineering series. And no, I'm not being payed to say that. It's totally free.
Well, the first question is, do you really need 10?
I actually did something similar to this to make a little flashlight circuit recently, using a 9v battery and a matrix of nine 3v 30mA LEDs connecting like so:
The thing to remember, here, is that connecting things in series means the current thru them is the same, but the voltage drops. Connecting things in parallel means the opposite -- voltage is the same, but current drops.
So in circuit 1, we see that we have 3 strips of 3 LEDs, for a total of nine. Each strip is connected in parallel, so the voltage across each strip is the same as the battery (9 volts). The current is triple what we'd have for a single strip of LEDs, but is still tiny, and well within what the battery can supply (~90 mA).
LEDs that function best at 3.2 volts can probably function at 3 volts, so this circuit should achieve close to maximum brightness with no power wasted on resistors.
Edit: It's notable that this design might have cascading failures over a number of possible things. If any one LED fails open circuit, one string will fail, and the current across the other 2 strings will increase. Conversely, if it fails closed circuit, it will put more voltage across the other two in the LED string, also potentially causing a cascading failure. This is all useful information, but this seems like an entirely acceptable product design as long as you consider what it means for the LEDs to fail. Your circuit should be prepared for the possibility that the entire cluster of LEDs becomes a dead short, and shouldn't catch anything on fire (fuses, anyone?).
Furthermore, if longevity and reliability is your priority, you might be better off using Transistor's schematic, baring in mind that you will waste a rather significant percentage of battery power. Or better yet, use a buck converter module. But all of these options, it seems to me, are out of the scope of OP's goal -- which was decidedly not to make a final product design for a company that needs to be UL listed.
If you need 10, this design doesn't work of course. One option is to do something like this:
Which one person already suggested. This would work, but generally speaking, the less voltage you have to drop across a resistor, the better. I'd recommend something more like this:
As you can see, we are dropping 6 volts across the LEDs, and are left with only an extra 3 to drop across the resistor. We know each LED takes 30 mA, so the total for each string of LEDs is also 30 mA (0.03 amps), since series of components have the same amps. Each string is connected in parallel, and we have 5 strings, so the expected total for the LED matrix is .15 mA. We can calculate the value of resistor to use, with ohm's law, like so:
Voltage (V) = Current (I) * Resistance (R)
V = IR
V/I = R
3 v / 0.15 amps = 20 ohms
Edit: I did a bit more calculating, and the resistor from this diagram, that I suggested you use dissipating .45 watts (0.15 * 3 = .45). Thus, I suggest you use at least a 1 watt resistor. There are also ways to use multiple smaller resistors as one larger one, to spread the heat out.
Edit 2: As Transistor pointed out in their answer, you can also split the resistor up into multiple resistors for each. The efficiency of this is exactly the same (about half a watt dissipated by the resistors in total), but it is actually considered somewhat better practice to split your heat across your circuit, rather than just using a bigger resistor. I mainly try to go for minimum component count in my designs, but you might be better going with Transistor's design if you want better longevity.
Addendum: as one person pointed out, Transistor's circuit also is more reliable if there are wildly varying electrical characteristics of the LEDs -- if one string pulls more current, the other ones will be affected, with my single-resistor circuit. So if you're going to use a resistor-mediated circuit, theirs is probably objectively best (in terms of reliability).