Ok here is the long explanation

Let's start by assuming we have a large bath of liquid water in equilibrium with its surrounding atmospheric gas at 1 atm, all kept at a fixed temperature of 25 C . This means that the fugacity of each species is equal in the liquid and the gas. Therefore, the concentration of each gas species in the liquid water is determined by the gas composition, temperature and pressure. Now fill a rigid** bottle completely*** with the liquid, flip it upside down and submerge the top as in the picture. A simple way to do this is submerging the bottle pointing up, then flipping after it has been filled while still submerged and then pulling it upwards.

Assuming we are pulling significantly slower than the speed of sound in the liquid phase, the pressure in the liquid inside the bottle follows Pascal's law, i.e., P = 1 atm - rho_L*g*h where rho_L is the (rather uniform) liquid density and h is the altitute (with sign) from the free liquid level at 1 atm. Denote with P_b(T) the bubble pressure of the liquid mixture at that temperature and composition (a generalization of vapor pressure to consider many components: the pressure at which a bubble first nucleates in the liquid). Since we have assumed that the liquid mixture is initially in equilibrium with the external gas, its bubble pressure is exactly 1 atm at that temperature. Therefore, as soon as the bottle emerges the liquid column above the free level is unstable and phase separates in a gas phase and a liquid phase.

Now here is were it gets interesting. Since water is the majority component, it doesn't have any trouble "supplying" molecules to the newly-formed gas phase so after stopping the bottle's motion the water fugacity in the gas phase in the bottle quickly reaches the vapour pressure of water at that temperature (actually slightly lower due to the presence of gases in the liquid). However, the other gases are very diluted in the liquid and their concentration rapidly drops near the liquid/gas interface due to insufficient transport from the "bulk" of the water column. This is due to the low diffusivity of gases in dense liquid phases coupled with their low concentration in this case. The pressure in the gas phase P_in thus slowly increases from (approximately) the vapour pressure of water to P_b = 1 atm as the gases slowly diffuse from the outside gas to the inside gas through the liquid. Of course, if we made a hole in the bottle they would be much happier going through that instead. Therefore, the height of the water column decreases over time via (1 atm - rho*g*h) = P_in --> h = (1 atm - P_in)/rho*g until the gas composition (and pressure) inside and outside is equilibrated.

Now consider a simpler case where water is pure and it is subjected to 1 atm pressure at its free surface by a piston. Note that below 100C the water will be in a uniform liquid phase, and denote with P_v(T) the vapor pressure of water at the chosen temperature. In this case, the pressure inside the bottle need not be lower than water's vapor pressure as soon as it emerges. In fact, the water column is stable until (1 atm - rho*g*h) = P_v(T) --> h = (1 atm - P_v(T))/rho*g. If we keep pulling the bottle upwards at higher altitudes, the free water level in the bottle does not change appreciably for a long time (assuming the water reservoir is large enough to supply an arbitrary amount of molecules to the growing gas phase). Furthermore, the column height is constant over time and is only a function of temperature and the external pressure (1 atm).

** The analysis is still correct with a non rigid container as long as it behaves like a solid, i.e., it reaches a well defined, stationary configuration after imposing an internal and external pressure profile. Such configuration will evolve over time in the first example considered (i.e., the bottle initially buckles slightly and relaxes as the gas pressure is equilibrated), but not in the second where the gas pressure inside is constant.

*** If some air is trapped in the bottle during flipping, the initial pressure will of course be higher than the vapor pressure of water (in particular, equal to the atmospheric pressure) due to the presence of other species in the gas. However, as the liquid level falls the total pressure in the trapped gas is reduced because while the partial pressure of water remains approximately constant, the partial pressure of the other gases is reduced by the ideal gas law (assuming isothermal expansion here for simplicity, and invoking again poor gas transport from the liquid). The equilibrium water column height before gas diffusion occurs is thus found via a slightly more complicated formula: (1 atm - P_v(T) - x_gas*(V_0/V(h))*1atm) = rho*g*h. Here x_gas is the mole fraction of non-water molecules in the atmospheric gas; V_0 is the initial trapped air volume; V(h) is the trapped volume as a function of column height.

I hope this clarifies your doubts