Electronic – Replace ONE 25v 47uf capacitor with FIVE 60v 10uf capacitors parallel

You are using your transistors incorrectly. In fact, I think that the only transistor which behaves as a switch in your circuit is T2.

Note that a transistor, NMOS for example, will conduct while \$V_{GS}>V_T\$. You are biasing your NMOS transistors in such a way that the potentials at source terminals is much higher than at gate terminals.

I believe that the reason you see a constant voltage on the load is because T3 and T4 are conducting when \$VG_2=5V\$, and T1 and T5 are conducting when \$VG_2=-5V\$ (in all cases due to wrong biasing). You can easily check my guesses in simulation by measuring the currents.

Furthermore, take a look at transistors' datasheets - the voltages in your circuit exceed the max ratings for the transistors.

The component you are trying to implement is called "Charge Pump" - there are many standard implementations you can find on the internet. A short (but not complete) overview may be found here.

Do the math.

Energy needed by the solenoid: \$12 \text{V}\times 500\text{mA} \times 2\text{s} = 12\text{J}\$.

You can charge the cap to 12 V and a boost switcher will convert the cap's decreasing output voltage to 12 V. Let's say the boost switcher is 80% efficient and can operate down to 2 V. So the cap needs to provide \$\frac{12 \text{J}}{80\%} = 15 \text{J}\$.

After discharging the cap from 12V to 2V there will be a fraction of the energy left: \$\frac{(2 \text{V})^2}{(12 \text{V})^2} = 0.028\$

So the cap must hold a total energy of \$\frac{15 \text{J}}{1 - 0.028} = 15.5 \text{J}\$ at 12 V.

The minimal capacity needed is: \$2\times \frac{15.5 \text{J}}{(12 \text{V})^2} = 215 \text{mF}\$

So, you need at least a 220 mF 16 V cap. This assumes you have a boost converter that can produce 12 V at 500 mA from 2-12 V and is at least 80% efficient.