What I don't understand is the math for how to figure out the exact voltage drop of the transistor between the collector and emitter.
You don't need an exact voltage. \$0.2V\$ is a reasonable estimate for most BJTs in saturation. The datasheet will give you more accurate values, under a range of operating conditions. \$0.2V\$ also isn't very significant to most circuits, so you can just ignore it. By ignoring it, you slightly reduce the current in the LED, which is erring on the side of caution, so isn't necessarily a bad thing.
And I'm also trying to figure out the math used to calculate the required milliamps that have to flow through the base of the transistor in order to fully turn it on (but not waste extra electricity).
There's a rule of thumb for a BJT used as a common-emitter switch, like this:
simulate this circuit – Schematic created using CircuitLab
when you want to drive the transistor into saturation (as you do here), make the base current 1/15th of the collector current. Again, the datasheet will give you more detail, but many of the parameters (like \$\beta\$ or \$h_{fe}\$) can vary over a wide range, as a function of temperature, operating current, and individual device manufacturing variation. The solution is to make sure you have plenty of base current so you are sure to saturate the transistor in all cases.
So:
$$ I_b = \frac{I_c}{15} = \frac{100mA}{15} = 6.7mA $$
The base resistor will have the \$5V\$ from the Arduino across it, less the \$0.65V\$ drop of the base-emitter diode across it, and the current is then given by Ohm's law:
$$ R_b = \frac{V_{R_b}}{I_b} = \frac{5V-0.65V}{6.7mA} = 652\Omega $$
Standard value of \$680\Omega\$ is close enough. The power in R1 is:
$$ P_{R1} = \frac{V^2}{R} = \frac{(5V-0.65V)^2}{680\Omega} = 0.028W $$
...so even a 1/8W resistor is fine here.
You mention that you don't want to waste electricity. There's not exactly much being wasted here; probably the current limiting resistor in series with your LED is wasting more electrical energy than this transistor arrangement. But, there are a few ways around it. One is to use a MOSFET instead of a BJT, which has the advantage of nearly 0 gate (equivalent to the base) current. 2N7000 is common and cheap and would do nicely here.
Or, you can arrange the transistor as an emitter-follower, so the base current goes towards powering the LED, and is thus not "wasted":
simulate this circuit
For more detail, see Why would one drive LEDs with a common emitter?
The big problem is that you didn't draw the schematic properly. A good schematic always makes the things clear:
simulate this circuit – Schematic created using CircuitLab
The current is \$I = \frac{U}{I}\$
I0 = I1+I2
I0 = (5V-2.065V)/270 = 10.87037037mA
I1 = (2.065V - 1.863V)/100 = 2.02mA
I2 = I0 - I1 = 8.85037037mA
Best Answer
The question, taking into account comments by the OP, seems to boil down to this:
LEDs are current-based devices, not voltage based. In other words, the design parameter for a driving circuit is, how much current should be allowed to pass through an LED. This current limiting is frequently done by putting a resistor in series with the LED (one for each LED if several are to be used in parallel).
The current rating of the LEDs is not stated in the question. Since many common LEDs are typically rated for 20 or 30 mA, assuming 30 mA (0.03 Amperes) for the purpose of this answer.
(this nominal value would apply for nominal current passing through the LED, and both values should be taken from the LED datasheet)
12.6 - 3.8 = 8.8 Volts
V = I x R
to the above voltage gives usR = 8.8 / 0.03 = 293.33 Ohms
The next higher commonly available resistor value of 330 Ohms would work well - Connecting one in series with each LED fed from the 12 Volt power rail should work fine. The higher than calculated resistor also increases the margin of safety by slightly reducing the current through the LED, without perceptibly reducing illumination.
However, there is another twist to this tale: The power rating of the resistors used.
P = V x I = V^2 / R
.P = 0.23467 Watts