To figure out why this happens, you need to think about what boiling is, and how it works.
As you would know, the water in the pot boils because its temperature was raised above the boiling point by the flame. This required a net transfer of heat from the flame, through the pot, to the water in the pot. Why did the heat flow in this direction? Because the flame is hotter than the water in the pot, even when the water starts boiling ($T_{flame} > T_{boil}$)
Now, think about the water in the bottle. The only way for it to get heat is through the water in the pot. As long as the temperature of the water in the pot, $T_{pot}$, is less than $T_{boil}$, it is still liquid, and it transfers some heat to the water in the bottle. The water in the pot boils off at $T_{boil}$, and can no longer transfer heat as efficiently to the water in the bottle.
This effectively means that the water in the bottle is restricted to a maximum temperature of slightly less than $T_{boil}$, and that is why it never boils.
Another way to think of this is, there must be a temperature difference for a heat transfer to take place. Since the maximum possible temperature of the pot water is $T_{boil}$, the temperature of the bottle water can never exceed this.
EDIT: Another factor to consider is the low conductivity of glass, which means a high temperature difference is required to let a small heat flux through.
What they were calling "sulphurous acid" back then is not what we would call an acid today. It was anhydrous sulphur dioxide which has a boiling point of $-10^\circ$C.
When liquid sulphur dioxide was poured into the red-hot vessel, due to the Leidenfrost effect, it would form itself up into globules and float on a layer of its own vapour. In this state the temperature of the globules would be just below that of its boiling of $-10^\circ$C as it evaporates away at a now greatly reduced rate. Pouring in a small amount of water, which freezes at $0^\circ$C, while the sulphur dioxide is in this state results in it freezing within a few seconds. Once all the sulphur dioxide has evaporated off, the ice will quickly melt again before being brought up to just below its boiling point of $100^\circ$C as it assumes its spheroidal form due to the Leidenfrost effect. If one is quick, before all the liquid sulphur dioxide has disappeared one can thrown out a small lump of ice from a red-hot crucible!
Update with a Reference
A source confirming the liquid sulphurous acid was indeed (anhydrous) sulphur dioxide can be found on page 645 of volume 3 of Lehrbuch der Physik by O. D. Chwolson, a copy of which can be found here. To quote, on the sixth and seventh lines down from the top of the page Chwolson writes (in German)
Boutigny brachte flüssige SO$_2$, welche bei $-10^\circ$C siedet, in einen glühenden Platintiegel, $\ldots$
which in English translation roughly reads
Boutigny brought liquid SO$_2$ [sulphur dioxde], which boils at $-10^\circ$C, into a glowing platinum crucible, $\ldots$
Best Answer
Evaporation is a different process to boiling. The first is a surface effect that can happen at any temperature, while the latter is a bulk transformation that only happens when the conditions are correct.
Technically the water is not turning into a gas, but random movement of the surface molecules allows some of them enough energy to escape from the surface into the air. The rate at which they leave the surface depends on a number of factors - for instance the temperature of both air and water, the humidity of the air, and the size of the surface exposed. When the bridge is 'steaming': the wood is marginally warmer than the air (due to the sun shine), the air is very humid (it has just been raining) and the water is spread out to expose a very large surface area. In fact, since the air is cooler and almost saturated with water, the molecules of water are almost immediately condensing into micro-droplets in the air - which is why you can see them.
BTW - As water vapour is a gas, it is completely transparent. If you can see it then it is steam, which consists of tiny water droplets (basically water vapour that has condensed). Consider a kettle boiling - the white plume only occurs a short distance above the spout. Below that it is water vapour, above it has cooled into steam. Steam disappears after a while, as it has evaporated once again.