Motorized Wheel Chair Source

Ok,

I'll give you the basic approach, but you would really need to tell me which arm of the steel angle is vertical and how exactly the angle is attached to the chassis (how many bolts, what's their spacing, what diameter, etc) since I don't know the chassis well enough to tell from your description.

Basically,
Stress = M/Z where M is bending moment and Z is the section modulus of the angle iron in the vertical direction. You can get a table of Z for various shapes from a steel supplier or calculate it (see a text book).

For the scooter, M = 0.8 Meters * 100kg Spread that over the two angles and you have a first cut.

Next, assume you need a minimum factor of safety of 4, so multiply the calculated stress by at least 4 to ensure the structure is strong enough.

Next, realize that the load on this carrier will be dynamic (ie: it bounces up and down as the truck does). This implies 2 things. First, you need to account for the additional load caused by accellerations ( assuming 1/2 G is probably reasonable). Second, the structure will be subject to fatigue. If the scooter is only carried occaisionally, fatigue will probably not be a problem, but if the scooter were to be carried every day for years, then you need to take fatigue into account. You can get an S/N curve for the steel you will be using to do this. (S = Stress, N = Number of cycles).

Assuming this structure is attached to the truck with 2 bolts through each angle iron, the moment is acting around the rearmost bolt and being resisted by the foremost bolt. The load on that bolt is the total moment calculated above divided by the distance between the bolts. The shear strenght of the bolt must be greater than this force (including the factor of safety). Also the single rearmost bolt must be capable of taking the shear due to the weight of the scooter.
To give you an idea, assume you were to use a 3/8 inch bolt.
Area = PI * r^2 = 3.24 * (3/16)^2 = 0.115 sq inches. The weight of the scooter is 100kg = 220 pounds. So the shear stress on the aftermost bolt would be 220 pounds/.115 sq inches = 1913 psi. With a factor of safety of 4 that would be a stress of 7652 psi. The yield strength of steel in shear is in the order of 11000 psi.

That's the basic approach. Make the long leg of the angle iron vertical, and have fun.

My gut instinct without running numbers is that the sizes you mention are in the right ball park. So do some calculations, and then take a look at some of the products being offered on the market. One thing you will notice is that a lot of the products out there will be based on square tubing rather than angles. Both square tubing and I beams are inherently stronger than an angle iron just due to their shape.