Your intuitive explanation is correct.
More formally, a transistor (the red NPN in that case) will start to transmit when the voltage on the base (in between the yellow resistor and the photoresistor) is above 0.7 volts.
The equation to calculate the base voltage is: Vb = Rphotoresistor / (photoresistor + Ryellow) * battery voltage
Like you predicted, if the photoresistor value become really small compared to Ryellow, the Vb will at some point fall bellow 0.7 volts and the transistor will be completely cut off.
That is a simplified version of what could really be going on. In reality, the base current will modulate the current that is passing through the Collector (the top of the transistor). The smaller the photoresistor, the more current flow through B and more importantly through C.
To conclude, the transistor can take different state: it can be cut-off (not letting any current flow between the collector (top) and the emitter (bottom), it can be in active mode (letting a current flow in between the collector and emitter proportional to the current flowing through the base) or it can be in saturation where the collector current for a given circuit is maxed out and an increase in the base current will not create an increase in the current circulating on the collector. The LED can be powered in the active or saturation mode. If the LED is powered in the active mode, chances are that you will see a dimming effect before a complete power off if the light change is slow. switching between saturation and cut off will act like an on/off switch.
To read further on transistors if you're interested: https://learn.sparkfun.com/tutorials/transistors
I hope this helps you getting started and I hope I was clear enough in my explanations.