1 watt second is a joule, so I will use J instead of W/sec
To be accurate you need to determine what is a definition of destroy. For example, to burn a person could mean incapacitate by 2nd or 3rd degrees burns of much of the body, or it could mean turn to ash.
Published data from atomic testing suggested that 8-10 cal / cm^2 is enough to cause 3rd degree burns on unprotected flesh. 10 cal / cm^2 is about 420 KJ / m^2. Average adult has about 2 sq meters of skin, but since you can only irradiate 1 side, you have to settle for burning one side of a person, so 400 KJ one the target will incapacitate a human. Without medical treatment 3rd degree burns over 50% of body is often fatal due to shock, fluid loss and infection. To actually guarantee a kill, you need more power on target of course, 10 MJ is probably enough and their clothes will catch on fire too overriding more common forms of protection outdoors.
Setting hardwoods on fire requires about 1 MJ / m^2 this would be sufficient for many homes, but a crushed white rock roof would be mostly unaffected, perhaps 10 MJ / m^2 would be enough -- 200 sq meters would include most homes so you need 2000 MJ or 2GJ on target.
A car again is difficult determine what destroyed requires. Unlike on TV cars to not catch on fire or explode at the least provocation. To temporarily incapacitate a car via space laser, the softest target is probably the tires, could not find direct data, but I suspect 1 MJ / m^2 would ignite the tires most of the time, but since the tires are often shielded by the vehicle they would often not catch on fire until far beyond this point and in any case you could easily fix this. To destroy a car you need to get it hot enough to destroy vital components like electrical wiring or engine belts - though again belts are easily replaced. I believe if you can heat the car to 500 C you will destroy most of the wiring as well as other soft components (seals, etc). So, how much energy to raise a car temp by 500 C (it could be winter). Curb weight on a new Ford Exposition is up to about 6000 lbs, as a first order approximation lets model this as 2720 kg of iron. Iron has a heat capacity of 0.45 J/g/deg. So 2720 kg * 1000 g/kg * 500 deg = 612 MJ on target.
Tanks are a lot like cars, so I will use the similar assumptions but I will require 800 deg and 63,000 kg (Leopold 2) and I will need about 22.7 GJ on target.
Office building. Likely brick, stone, steel in construction. Again how much is enough to consider it destroyed? If you really want to be sure, you need to err on the high side. The Empire State building used 60,000 tons of steel, 200,000 cubic feet of limestone and granite, 10 million bricks and 730 tons of aluminum and stainless steel. To simplify, I will assume 200,000 cu ft of limestone and ignore the bricks and aluminum. How much does granite weigh about 2.75 g/cm^3 and a heat capacity of 0.19 J/g/deg. Total granite mass 15,600 long tones. For a 500 deg temperature rise I need 13,500 GJ for the steel plus 1,482 GJ for the granite or about 15 TJ to take out the empire state building. In actually fact, you don't heat up the lower floors in this case as the upper floors absorb nearly all of the heat. A 1 second pulse would need to essentially vaporize the upper part of the building in order to destroy. But that is not even enough as the vapor cloud itself would absorb and scatter a large percentage of the laser blast. A thousand times might be enough though.
A small town, again what is threshold of destruction, and just how big is this. I was raised in Columbus Indiana, so seems like a good model to me. Columbus IN, population 44,061 (2010 census). Area 72.23 sq km. 500 deg temperature rise and steel & concrete destruction (to take out all of the buildings). I will simply revert to the energy per unit assuming the 100 MJ per sq. M is enough to take out everything (100 times the energy to ignite hardwood). 100,000,000 MJ / m^2 * 72,230,000 m^2 = 7.223 PJ -- petajoules on target.
Needless to say, there are no experimental confirmations of the beam attenuation that would be encountered firing from orbit to surface targets, certainly not anything approaching these energy levels. However, I did find Laser Atmospheric Attenuation Tables for LTAS. Without going through the justification at this I think you would need about 20-50 times as much light at the source to reliably punch through the atmosphere as you have at the target. I have not yet performed more detailed calculations to confirm this. So, take all of the amounts above and multiply by 50 if you want to be able to destroy targets with reasonable certainty. This is based on bad weather conditions. For best case but still realistic weather conditions it would appear that you only need 5-10 times the on-target energy at the source.
The higher energy levels deserve better accuracy as their are many non-linear and probably classified effects that doubtless come into play.
I am also understating the actual power required because I generally ignored that fact that for everything except the unprotected human case, the target will ablate. The surface layers will cook off turning into gas and taking much of the incoming heat with them, and continuing to absorb and scatter the incoming laser light. I also generally ignore the fact that the target will reflect some of the laser light, when in can reflect at least some of the incoming light. At most of these energy level, a mirror will not provide protection, at least not for long as even a good mirror will absorb enough energy to quickly be destroyed.
These power levels are beyond ridiculous. For everything larger than a car, the power at the target exceeds that of even the largest electrical generation stations. The entire United States only has a electrical capacity of 1060 GW.
For our simple 3rd degree burns on a single human 1MJ on target requires a minimum of 5 times that at the source and 5 times again for the loss of electrical conversion to laser energy, i.e., 25 MJ. A common comparison is that 1 MW will power 1000 homes, so 25 MW is over half of the power needed to supply Columbus Indiana just to incapacitate 1 person.
Using capacitors to store up for a bolt from the sky may sound like a good idea, but your requirements dwarf standard capacitors. The largest capacitor bank in the world is only 50 MJ, built at a cost to 10 million Euro
Somewhere during your construction phase people are going to wonder what you are building and will get motivated to put a stop to your evil plan.