Let's look a few things realisticly:
One operating a power tool from an inverter from a car is an isolated power source. Ground is insulated by the tires of the car. The likeliness of getting shocked is almost nil.
The next part of reality is the 1800 W is a peak power and not what it would draw at all times.
The next part of reality concerns the lighter socket. This is fused at probably 15 Amps an probably has some other stuff attached to it. Since P=VI then the max power that can be taken from the lighter is about 12 V * 15 A which equals about 180 W. Then there is an efficiency factor of maybe 0.8 so therefore it's even less. 0.8 * 180 W. So, the lighter socket won't cut it.
The inverter must be connected to the battery directly. To look at the requirements of 1800W say at 90% efficient. This is about 2000 W. Again using P=VI we find that the current needed at 12V is 2000 W/12 V or about
166 AMPS.
Yes 166 Amps. The battery can deliver this for short periods.
Let's say the car has a 60 Amp alternator, so therfore you can deliver 60 A * 12 V or 720 Watts continuously.
Again remember that you are not operating the chain saw continuously at full load.
Here is just one example of a 2000 Watt inverter:
Vector 2000 Watt Power Inverter VEC054D
It would work for you.
There are basically 3 types of inverters:
1) square wave - not recommended for anything
2) Modified sine wave
3) Sine wave- The most expensive and EXPENSIVE.
The power that you require will need the inverter to be connected to the battery and the car would have to be running.
Direct battery connection can be done with clips. Side terminal batteries can be a problem. Car stereo installers have available ways of tapping the battery connections with very high current fuses.
The battery connections must be made with the inverter off.
I have an electric chainsaw, but I use it rarely or trimming large tree branches. Usually no more than 4" in diameter.
Does everything make sense?