It is not clear from your question whether the gas in the bottle starts out as air, or the only substance in the bottle is water, either in liquid or gas form. I'll assume the latter since it's easier to answer.
Once the bottle is closed, ambient temperature doesn't matter. The pressure of the gas part in the bottle will be stricly a function of temperature, with that function being solely a property of water. You can find what that function is in what is commonly referred to as a steam table.
As you heat the bottle, and presumably everything inside gets to the same temperature, the vapor pressure increases. This causes a little more of the liquid to boil and thereby make less liquid and more gas. Eventually for any one temperature, a new equilibrium is reached where the gas is at the pressure listed in the steam table for that temperature.
If you let the system reach equillibrium at 200°C, for example, then open a valve at the bottom, water at the bottom will be forced out under pressure. As the volume of water in the bottle decreases, the volume available to the gas increases, which decreases its pressure. The liquid that was at equillibrium now no longer is, since its pressure is reduced but its temperature (for the short term) remains the same. This will cause the liquid to boil and make more gas. This cools the liquid, and if done slowly enough the liquid will track the ever decreasing boiling temperature for the ever decreasing pressure.
Meanwhile, the gas is expanding, so it will cool. It won't condense because there is no place for the heat to go. We are making the assumption that the bottle is insulated at this point. Boiling liquid will keep the gas "topped off" at whatever temperature it is according to the steam table.
Once all the liquid is gone, the steam will still be at pressure above ambient. If the valve stays open, the steam will vent until its pressure is the same as ambient. If the bottle is truly insulated, nothing more should happen. If heat is lost thru the bottle, then the temperature will decrease and the pressure of the steam decrease, again according to the steam table. Outside air is then sucked into the bottle. If this air is at 20°C, for example, it will significantly cool the steam, which will cause much of it to condense, which creates lower pressure, which sucks in even more cool air, etc. The proceeds until most of the water is condensed to liquid, with the only remaining water gas being whatever maximum partial pressure of water that air at that temperature and pressure can support.
If the valve is opened so that the 200°C water is expelled "quickly", then it gets a lot more complicated because there is not enough time for the system to track equillibrium as pressure is abruptly release. In that case, I think all we can say is that a bunch of super-heated water will be violently released and quickly undergo decompression, which will cause some of it to boil quickly, making a lot of steam, with the left over liquid being at boiling temperature for ambient pressure, which would be 100°C. After the water is gone from the bottle, a rush of steam will come out, condensing in the (relatively) cold ambient air and adding to the already large saturated water vapor cloud. Once enough steam is expelled so that ambient pressure is reached the remaining steam acts as above.
Adding salt to water makes it freeze at a lower temperature. This fact is being used in two different ways in the two scenarios you mention. Dissolving sodium chloride in water is slighly endothermic, but this effect is small and to the best of my knowledge isn't important in the drink cooling process.
Putting salt on the highway is quite straightforward: we don't want ice to form, so we put salt in the water to prevent that. This doesn't just change the amount of time it takes ice to form, it actually completely prevents ice from forming, unless the temperature gets so low that the water can freeze even with salt in it.
Cooling your drink is a bit more complicated, because in this case the rate at which things happen is important. You don't want your drink to be less than $0^\circ C$ because it would freeze; instead you want to cool it down to a few degrees Celsius nice and quickly.
The rate at which it cools depends on two things: the temperature of its surroundings (the colder the better) and the heat conductivity between it and them. You could try to cool it by putting it in a bowl of ice at $0^\circ C$, but the problem is that the ice is solid and will only touch the bottle at a few points. This results in a poor thermal conductivity, so the drink will only cool slowly.
To get around this, you could try mixing the ice with some water. Now the bottle is touching the liquid over a large surface area, and the liquid itself has a higher thermal conductivity than solid ice due to mixing, so heat will be transferred much more quickly. But the problem is that the water won't be at zero degrees any more, at least not at first (I'm assuming the water comes from a tap, so it's not chilled initially). You have to wait for quite a bit of the ice to melt before the water's temperature will drop. Also, once you put your warm drink into the water it will heat the water up as the drink cools down, so again you have to wait for the ice to melt in order for the water to cool again.
The solution to this is to make the ice melt faster. You can do this by adding salt. This lowers the freezing point, making the water less "happy" about being in the liquid state, so it melts more quickly. This means firstly that the tap water you've added to the ice will cool to close to $0^\circ C$ much more quickly, and secondly that once you've put your drink in there the water will stay cold as the ice continues to melt.
It's also possible that, with the salt added, the water can go to below $0^\circ C$, but this will only happen if the ice is quite a bit colder than $0^\circ C$. This could be the case, but my intuition is that the rate of cooling due to the ice melting faster is more important here than the final temperature. You could easily test this by putting a thermometer in the salty ice water and seeing if it goes much below freezing.
There's also the fact that dissolving the salt is endothermic, as you mentioned. To test whether this is important, you could try adding salt to some chilled water without any ice, and see if the temperature drops a lot. My feeling is that it will only drop by a tiny amount that will be hard to measure with a normal kitchen thermometer, but you can always try the experiment.
Best Answer
When two phases are in equilibrium the chemical potential of the atoms/molecules in the two phases are the same. If you're not familiar with the concept of chemical potential it basically just means that the molar Gibbs free energies of the two phases are equal so the $\Delta G$ for the phase change is zero.
The argument is:
If the liquid and vapour are in equilibrium then the chemical potential in the liquid $\mu_l$ and vapour $\mu_v$ must be the same: $\mu_l = \mu_v$.
If the solid is also in equilibrium with the vapour then the chemical potential of the solid $\mu_s$ and vapour $\mu_v$ are also the same: $\mu_s = \mu_v$.
And that means that the chemical potential of the liquid and solid must be the same, because both are the same as the vapour: $\mu_s = \mu_l = \mu_v$.
And finally if the chemical potential of the liquid and solid are the same then it means the liquid and solid are in equilibrium i.e. we are at the freezing/melting point.
Re the second paragraph: when the solvent freezes it will freeze to form pure solid solvent i.e. it excludes the solute. So we have pure solid solvent in equilibrium with the vapour and "solvent + solute" in equilbrium with the vapour. So it's the same argument as above.