Yes, the "diode" is the LED.
There is no such thing as "wide open"- the current at the LED is reflected (within limits) at the transistor by the ratio "CTR" = Current Transfer Ratio.
If you put 5 mA through the LED you get somewhere between 2.5mA and 20mA through the transistor (until it saturates)- that's what the minimum/maximum CTR figures of 50% to 400% at If = 5mA and Vce = 5V mean. Vce =5V means that it's far from saturation. So if you use a resistor in series that limits the current to (say) 1mA (eg. 5K on a 5V supply) you'll have it saturate with 5mA into the LED. Note the since they vary over an 8:1 range, the manufacturer has ranked some of them and marked them in different "bins"
N : 50 to 400 (%)
H : 80 to 160 (%)
W : 130 to 260 (%)
Q : 100 to 200 (%)
L : 200 to 400 (%)
Naturally, the N version will tend to be the cheapest since the CTR can vary over the widest range, and includes the worst-performing units (50-80% CTR).
The transistor in the NEC part is rated at 70V and you should not put more than that across the Emitter-collector. There are higher voltage rated solutions.. for example the Sharp PC851XNNIP0F is rated at 350V.
You should make sure you have plenty of CTR- it degrades with temperature and with time (as the internal LED fades). Putting extremely high currents (like 30mA) through the LED will hasten the deterioration.
A proposed model of optocoupler aging is life \$\propto \frac{1}{Ie^{\frac {-E}{kTj}}}\$
Where k is Boltzman's contant 8.62\$\times 10^{-5} eV/K\$
Tj is junction temperature
E is the activation energy of approximately 0.15eV
so if you increase the current you not only get a decrease due to the current itself but an exponential decrease due to self heating. If I plug some plausible numbers into that equation, I get more than a 1000:1 reduction in life at 20mA vs. 5mA, with the same ambient temperature.
Note that if a relatively high current is only seen with a very short duty cycle (perhaps fitting your application) then the life is hardly impacted. It's the current and temperature integrated over time that causes the deterioration.
Bottom line is that you want to keep the current as low as is reasonable to make the thing work (and since operation is guaranteed at 5mA, that's not a bad place to start). You should also make sure it will work at (say) 3mA so even if it ages a bit it will still continue to work. In some cases you may have to buy more expensive optos with higher CTR than the cheapest ones if you need long life. Anecdotally, there is significant difference between different manufacturers' products. Personally, I tend to stick with the best-known Japanese makers.
Best Answer
To start with your last question: yes, it's a microcontroller (not a microchip, that's a confusing word since it's the name of a manufacturer of a.o. microcontrollers).
Unfortunately the supply current is not for you to choose. At 3.3 V it will be typically 0.35 mA, but it can be anything between 0.27 mA and 0.45 mA. And the 10 kΩ you calculated is the equivalent resistor of the receiver's power consumption.
The resistor in the schematic forms a low-pass filter with the capacitor, letting DC and low frequencies pass and filter out the higher frequency noise. That resistor will cause a small voltage drop from the power supply; the higher the current, the higher the voltage drop, due to Ohm's Law. So worst case the current will be 0.45 mA, and then the 1kΩ resistor will cause a 450 mV drop. If you supply 3.3 V then the receiver will have 2.85 V left. This is enough to operate, but you have to take into account that it will only output 2.85 V as a high level, and you have to check that this is enough for the microcontroller.
You're right about the working of a decoupling capacitor, and that's part of its job. Like I said the capacitor is also part of a filter. If you choose 1 kΩ for the resistor, and 100 nF for the capacitor it will have a cutoff frequency of 1600 Hz.