Hi curiouser,
the circuit is about right (don't have the 6N137 data sheet at hand presently). It should be wired the way its output is high when the receiver LED is under current. With the high output on the base of the 2N2222 it conducts and switches on the IR-emitters. Base resistor should be 1 to 2K7.
There are current limiting resistors necessary for each IR-emitter, but you might connect two times two in series. Using the TSAL6100 it is a real powerful IR-emitter with an IPCE of >100mW/sr (normal IR-emitters are at 25 to 50mW/sr)
The forward voltage is 1.3V(Uf) and the forward current (If)is 100mA.
Two examples of wiring the emitters: Two parallel pairs in series and all four parallel.
Current limiting resistors: two pairs: (5V-(2*Uf)/If), (5V-2.6V)/0.1 = 24Ω, that is two 24Ω resistors. (preferrable method)
All four parallel: (5-Uf)/If, (5 -1.3)/01 = 37Ω, that is four 37Ω resistors.
Paralleling four IR-emitters the transistor has to sink 400mA total, using two pairs of two the current is 200mA total.
Don't be worried about the 38KHz of the emitters. The MCUs frequency is much higher.
Don't forget to use a current limiting resistor for the opto-isolator as well, if none - ZAP!
Boncuk