The DC case is very simple - and your understanding will do for the moment.
The AC case is not that much different .... if you take things s l o w l y.....
Consider a 4 step process, where one connection to the capacitor is taken as our reference point and the other is the one we are monitoring...
1. Charge up a capacitor with a DC source of positive polarity that takes 10 seconds for the capacitor to reach the DC voltage
2. Discharge the capacitor with a load that takes 10 seconds for it to effectively have zero volts.
3. Charge up a capacitor with a DC source of negative polarity that takes 10 seconds for the capacitor to reach the DC voltage
4. Discharge the capacitor with a load that takes 10 seconds for it to effectively have zero volts.
This is 4 steps involving DC - but put them together and you have AC .... and if the AC frequency, supply and capacitance were to produce the same time constants, then the observed characteristics would be comparable. (Yes, the 'signal' would be an unusual sort of square wave.)
But now lets look a little bit more closely....
In the very first instant of step 1, a DC supply is connected to a capacitor with no voltage across it. A capacitor CANNOT change it's voltage instantaneously - (and here is the important bit:) it needs to build up that charge over time. To do this, current needs to flow. At the start, this current flows quickly limited by the resistance in the system, then as the charge builds up across the capacitor, the voltage increases - and increases more and more slowly until it eventually reaches equilibrium with the supply. At that point, current flow stops.
Over the next 3 steps, the exact same process occurs - just with different voltage levels.
Now, at 10 seconds per 'stage' this exercise is somewhat straightforward and a bit boring ... but let's speed things up a bit - say 1 second per step, but using the exact same capacitor and supply.
At step 1, we start to charge up the capacitor. A big current flows because of the voltage difference, but there is no time for it to climb up to the supply voltage before we cut it off and start step 2.
At step 2, we discharge.
At step 3, we repeat the charging in the opposite direction - with the same cutoff.
At step 4, we discharge.
Now let's peed things up even more - say to 1/100th of a second per stage (25Hz)
At stage 1, the current is still high when it gets switched off and the voltage hasn't had time to increase much at all.
Stages 2, 3 and 4 follow the same pattern as before.
Now let's speed things up even more - say 1kHz
The current is essentially at it's peak value for all of the short time it's on, only changing from positive to negative and back. The voltage across the capacitor does not have any time at all to change by any measurable amount. Notice that the higher the frequency, the greater this phenomenon becomes. (This is important - and the relationship between frequency, capacitance and reactance is one you will see a lot more of...)
Now we have a fair idea about what is happening inside the capacitor, let's see how it fits into a couple of situations...
If you were to have this capacitor across the power supply, then it would - in effect - short it out. The AC part of it, that is. The DC is left behind.
However, if you were to have this capacitor in series with a signal, then the voltage difference between both sides will be essentially unchanged, since there won't be time for charge to build up in one direction before that direction is reversed. Once you've got a grasp on this, we can go one step further and look at a DC offset.
Edit: Change some wording for better accuracy.