The circuit you show should turn on the output LED when you interrupt the beam. But you also have to prevent any ambient light from being seen by the photo-transistor.To do that you may have to add a tube around the transistor and/or add a dark red filter.
You have over 60mA going through the IR LED which seems like a lot. Is that how much you want?
Honestly, i am sorry but i am not good at the technical details.
In the end, it will not be 60mA because there will be other, more complicated, sections added to this foundation.
All i need right now is to make it so that when the beam is interrupted, an output is high, as opposed to a constant high output, which i am getting right now
If your 9V is a brand new 9V battery then the current in the 1.3V IR LED is (9.0V - 1.3V)/270 ohms= 28.5mA which might be high enough to burn it out. If your IR LED survives such a high current then the phototransistor should be turned on shorting the red LED turning it off.
When the IR beam is interrupted then the red LED turns on and has a voltage of about 1.8V.