You are on the right track. You need a bistable circuit, which halvens the frequency of the photocell signal, and gives high output every other pulse.
Besides that, you may want to use a monostable multivibrator to adjust the duty cycle (i.e. the "high" time), in order to get pulses of the right length.
The thing you seem to be missing is edge triggered flip-flops. These are used at various points between combinatorial logic. They essentially sample the output of the logic at a clock edge, then freeze that value until the next clock edge. Meanwhile, the combinatorial logic is free to produce garbage intermeiate values, as long as it settles to the next answer before the next clock edge.
This is one reaon why most processors have multiple clock edges per instruction "cycle".
In your example of a 1-bit memory having its output fed back to its input via a inverter, there would be a edge-triggered flip flop in there somewhere. This would likely be either on the input or the output of the memory cell, often in both places. Even with just one flip-flip in the loop, the value is frozen for each clock cycle. By the end of that clock cycle, the opposite of that value is ready at the input of the flip-flop. At the next clock edge, the flip-flop samples its input, transfers that value to its output, and freezes the output again. In this scenario, you'd end up with the digital value toggled every clock cycle. In effect, you'd get a square wave at half the clock frequency.
With multiple flip-flops in the loop, each flip flop would add 1 to the overall divide value. With two flip-flop, you'd get the clock frequency / 3, with three you'd get clock frequency / 4, etc.
Best Answer
An HC chip needs >70% Vcc for the high voltage, which an LS gate is not guaranteed to deliver.
Use a pullup resistor on the output of the LS. Its output is rated to sink 8mA, so you can go down as far as a 620ohm resistor, though a higher value will use less current in the low state. 1k is a reasonable value to try. You cannot go too high in value, otherwise the pullup speed into load capacitance will be too slow and still not fix your multiple clock transitions. You will need to experiment with this if you want to use less current.