Yeah, it doesn't matter where the evaporation occurs. If it evaporates off of the outer surface, having seeped through the ceramic, then it cools the ceramic and the liquid water left behind. If it evaporates from the surface inside, then it cools the water inside directly. When a water molecule changes state from liquid to gas, it removes heat from whatever is in contact with it in order to make the phase change. The whole system reaches a thermal equilibrium over time. Note that the temperature depression achieved is a function of the *rate* of evaporation. The evaporation process sucks up 2260 J/g of water evaporated. So multiplying this by the amount of water evaporated per unit time gives one energy/time units (J/sec, for example). Energy/time is called "power". So as the water evaporates, there is energy being transferred FROM the remaining liquid water TO the ambient air (which is where the water vapor goes after it evaporates). This produces cooling. However, the ambient air, being a higher temperature than the remaining water, is exerting a heating affect on the remaining water by convective heat transfer, and also by conduction from the countertop into the base of the jug. So, then, if the evaporative cooling power is greater than the convective/conductive heat transfer back into the jar, then cooling will happen. The convective and conductive heat transfer back into the water is a function of the temperature difference -- the higher the "delta T" the faster the heat transfer. So what will happen is that evaporation will cause net cooling power up to a certain delta T. If you can slow that heat transfer back into the water from the environment, you can increase this equilibrium delta T, which is why the jug is made from an insulating substance like clay. If the jug were made of aluminum which would readily conduct the ambient heat of the countertop back into the water very easily, no net heat transfer would be established.

From all this we learn: if you stick this jar in front of a fan, you should be able to achieve a greater delta-T. And also, if you put the base onto a piece of styrofoam, you can probably squeeze out even a little more delta-T (by reducing conductive heat transfer from the base). Finally, the rate of evaporation is a function of surface area, so the fuller this jar is, the more damp clay there will be, and the greater the cooling power. It would be fun to use a thermocouple and see what kind of delta-T you can get for different conditions (with/without insulating base, with/without forced airflow, full/half-full). Ambient temperature is also a factor, but I can't remember to what degree (linear or not).

"Partial pressure" is the way of talking about the fraction of water vapor in the air. If the air has been dried, then it has no water in it and we say the "partial pressure of water vapor is zero" (in other words, none of the absolute pressure of the mixed gas called "air" is being caused by the presence of water vapor). Now if you put that dry air next to some liquid water, some of that water is going to evaporate into the air until an energy equilibrium is reached. The humidity where this equilibrium occurs is dependent on both temperature and pressure, with warmer air capable of holding more water vapor.