You are making this way too complicated. You don't need a opamp and a logic gate to perform your very simple logic. The on/off switch can simply be wired in series with the supply and act like a normal on/off switch. The inversion you want from the momentary button can be accomplished by a single transistor:
Q1 acts like a low side switch, with R2 providing it's base current and thereby keeping it on. When SW1 is pressed, it shorts the base to ground, which turns off Q1. The LED and buzzer are wired in parallel, not in series as you had them. It is unlikely both need the same amount of current. The LED will have about 10mA thru it when on, which lights a ordinary LED bright enough for most purposes.
Nearly the full 12V will be applied to the buzzer. You didn't say what voltage the buzzer was rated at, so I picked 12V just for example. That's a common voltage for buzzers, but they are certainly available in other voltages too. Some buzzers can appear inductive from the outside circuit point of view, so D2 protects the circuit from the inductive kickback when the buzzer is turned off. Buzzers can also make a lot of high frequency electrical noise. C1 is there to shunt most of that locally to reduce the radio interference and other bad effects from this.
The answer given by @AaronD is pointing completely in the right direction, but it's leading you further than you need to go (the combinational logic), and it's not entirely telling you how to get there.
The suggested article on Rod Elliot's website (sound-au.com) is a great starting point, but doesn't quite cover, or truly explain what you need.
Pinching the core of 'Project 144' - although it's a well established principle - is considering a resistor ladder. It's a string of resistors in series, with one end connected to a voltage source (we'll call it Vin) and the opposite end connected to ground. Each junction between any two resistors will, by Ohm's law, be at a specific voltage between Vin and ground. I'm not going to explain the maths behind it - that's your job to find out.
I assume you know that op-amps can be used as comparators - again I'm not going to explain this. You connect each junction of the resistor ladder to the reference pin of a single comparator, and a voltage between Vin and ground (let's call this Vtr) to all of the comparators. As you should have figured out by now, as you increase/decrease Vtr the comparators will switch state one by one.
Thus:
1) You need to create a voltage source (Vtr) that increases or decreases with time, with a fairly linear response for 9 seconds - that's how long your outputs need to run for. A simple RC circuit will do. It's your choice which 'direction' you choose, although if I were doing it I'd pick a decreasing voltage; the rest of my answer assumes this.
2) You need to build a resistor ladder, connected to your comparators. How many op-amps do you need? Well, you have a green light, a red light, and a buzzer; so three. That means you need four resistors. You need to pick values of resistors such that the junction between Ra (the topmost) and Rb (the second one) is the same voltage that your Vtr will have reach after your '2-4 seconds' required for the green LED to be lit. The next junction between Rb and Rc has to be the same voltage that Vtr will be after another '2-4 seconds' to cover the red LED. Guess what's next? Yes, the buzzer.
simulate this circuit – Schematic created using CircuitLab
The above schematic explains a little better the two points. This circuit is for a decreasing Vtr; if you want an increasing Vtr you'll have to change the order and orientation of the op-amps.
But what do you think happens when you connect the LEDs and the buzzer?
You should notice that when Vtr is equal to Vin, ALL the outputs are on! Of course they are, because all the comparators are being triggered. If you can't use anything other than op-amps then you can't use logic gates.
I don't know if you'll find this answer by Googling it, so I suppose it's not against the 'Stack Exchange will not do your homework' policy to give you the answer straight. Set the outputs of the comparators like this:
simulate this circuit
NOTE: I haven't included any dropping resistors in the above schematic - again that's up to you to calculate and include.
You should be able to work out what's going on and how/why it works. If you get a bit stuck, draw a truth table showing each comparator output ('hi' or 'lo') against time:
| Time | Vsup | Comp1 | Comp2 | Comp3 |
=================================================
| 0-3 | HI | LO | LO | LO | <-- GREEN ON
| 3-6 | ? | ? | ? | ? | <-- RED ON
| 6-7.5 | ? | ? | ? | ? | <-- BUZZER ON
| 7.5+ | ? | ? | ? | ? | <-- ALL OFF
The 'Time' column is of course not precise, it's just an indication of the four stages of output. Fill in the gaps and you'll see why only one output is active at a time.
If you have questions please comment, but make sure you show that you've put some effort in to get further than you are now.
Best Answer
First basic error: you are driving the OPAMP inputs with single resistors connected to the supply lines: this simply ties the inputs to the same voltages, as there is virtually no current flowing in them.
So, first thing to do is understand what is a voltage divider.
Then, you have to use the principle of the voltage divider to create voltages to the inputs of the OPAMP. Of these voltages, one will be fixed, and will be used as a reference, while the other will be variable in function of the value of the photoresistor.
The photoresistor
Let's talk about the photoresistor: this is a resistor that decrease its value when exposed to light; let's consider it in a binary way (for your purpose): low when there is light, high when it's dark. Now you have to specify what do you want as 'light', and what do you want as 'dark'. Then measure the value of the resistor in the two cases. If you are simulating, I don't know how to create these situation, try to take a look or use a variable resistance to simulate the effect.
Depending on the values chosen for the photoresistor, you will have to size the other 3 resistors such as they will generate a positive differential voltage in one case, and a negative in the other.
The AND gate
There is a "small" problem also with the AND inputs: the switch input, when this is open, will be floating and with an unpredictable value; so you should put a (quite big - 10K should work, for sure in the simulation) resistor, between that input of the AND gate and ground. This will pull down the voltage when the switch is open; but you really need the 680 Ohm? For sure not in the simulation, you can think about it in a real implementation.
(Sorry but I'm busy, I'm adding the answer a piece at a time...but try to understand those parts)