Well, the argument is not very well put:
Because the air is a bad conductor of heat the soup stays hot longer: only the first layer of air touching the soup gets hot fast, and heat is not transmitted efficiently to the bulk of air.
For soup, in contrast to thermos, evaporation cooling should also be considered.
Convection by continuously replacing the contacting layer of air increases the heat transfer to the bulk of air by conduction and at the same time the rate of evaporation increases, increasing cooling.
So the soup cools faster than if there were no convection.
If air were a good conductor of heat, the soup would cool fast, as in a metal plate on a metal surface.
You could go through a read of the wiki article.
Edit: Georg's comment makes me add that of course the soup would be also cooling because it will be radiating with the corresponding to its temperature black body spectrum. Convection increases the rate of heat loss over the loss through radiation.
To address the title, which differs from the questions in the content:
Why is air a poor conductor of heat?
It is mainly the very low density of gases that make them bad conductors of heat. In liquids and solids atoms and molecules are densely packed and transfer of energy has much smaller distances to happen. In a gas molecules have to scatter off molecules to exchange energy in larger distances, so the probability of transfer is much smaller.
There isn't a simple answer to your question.
Suppose you had a solid bottle rather than one containing liquid water, then you can solve the heat equation to describe the change in temperature with time. You'd probably need to do this numerically as only special cases like spheres would allow an analytical solution. However even then you can't answer a question like "how long does it take to heat to xx degrees" because the object isn't at a uniform temperature. There would be a temperature gradient from the surface to the middle.
With a bottle of water life is even more complicated because the water inside the bottle will develop convection currents as it starts to heat and these make your calculation even more difficult.
However you can probably simplify the problem if you assume that the water in the bottle is vigorously stirred and the water outside is vigorously stirred so that all the water is the same temperature. In that case the heat flow will simply be controlled by the thickness of the bottle walls and the thermal conductivity of the bottle walls.
If we simplify the system slightly and assume the heat flow through the bottle wall to be one dimensional then we can use Fourier's law for heat flow:
$$ \dot{q} = k \frac{T_w - T_b}{d} $$
where $T_w$ is the temperature of the water bath, $T_b$ is the temperature of the water in the bottle, $k$ is the thermal conductivity of the bottle wall and $d$ is the thickness of the bottle wall. To solve this we take the external temperature $T_w$ to be constant, and note that the change in the bottle temperature $T_b$ is equal to the heat transfered divided by the total specific heat of the bottle contents. Without going through all the details, solving this equation gives:
$$ T_w - T_b = A \space e^{-Bt} $$
where $A$ and $B$ are constants that contain contributions from things like the thermal conductivity of the bottle wall, the wall area, the wall thickness and the heat capacity of the fluid in the bottle. I've rolled everything into the two constants because to be honest calculating from first principles is hard and we'd usually just do a few experiments to calibrate the system. Once you've measured the constants you can use them to calculate the behaviour of the system for any initial conditions.
Best Answer
It would be more scientific to have compared a covered with an identical non-covered bowl, with temperature measurements over an identical period of cooling time.
But covering a bowl of hot soup with a plate does indeed reduce heat losses and slows down cooling of the soup, because:
Radiative losses are usually small at these temperatures but the plate further reduces them by partial reflection of the radiation. White plates will do that better than dark coloured ones in that respect.
Convective losses too will be reduced because convection relies on air movement which becomes somewhat restricted by the plate.
Evaporative losses become small once the head space between soup and plate has become saturated with water vapour.
Only convective/conductive losses from the body of the bowl will not be affected by the plate cover.