Decoupling and coupling are different things.
Have a read of my long posting in this thread:
https://www.electronicspoint.com/newbie-questions-t248766.html
Ignore the part from the LED onwards. I think some people didn't like it because of the strange analogy I used
In it, I model a capacitor as one of those hydraulic suppressor contraptions they attach to the top of doors, to stop them from banging closed. They will move, but only at a certain speed. The more pressure (current) you apply, the faster the plunger will move in the suppressor (the faster the voltage across the capacitor will change). If you hold the door open and push and pull it quickly, the rapid changes in the force on the door are transferred to changes in force on the wall where the suppressor mounts, but the suppressor itself doesn't compress or expand much in response to quick changes in force. It's the same with alternating current and a capacitor. The capacitor can slowly charge up to the DC voltage you apply (i.e. the suppressor extends when you open the door slowly), but brief changes don't cause its length (the voltage across it) to change much.
This analogy applies best to larger-value capacitors, e.g. say 100 uF and higher, where the charge and discharge times can be in seconds like a door suppressor. With small value capacitors the effect is there, but less force is required and/or the suppressor moves a lot quicker for a given amount of force. If you include ALL types of them, capacitors as a group cover a capacitance range from under 1 pF (picofarad, 1x10^-12 farads) to over a farad, a range of more than twelve orders of magnitude.
You can use a capacitor in two basic ways. Either IN SERIES with the signal (coupling) or IN PARALLEL with the signal (decoupling, or "coupling to ground").
Decoupling is the simplest to understand, and the door analogy applies here. A capacitor to ground (ground is an unmovable reference voltage - most of the time) is equivalent to a hydraulic suppressor attached to a wall (an unmovable reference point). Anything that tries to open or close the door quickly meets "resistance" (not resistance in the electrical sense - conductance is the electrical equivalent, which is the reciprocal of resistance), so rapid changes in the door position are reduced. This is how decoupling works. A voltage rail that is decoupled with a capacitor to ground is held at a relatively steady voltage, regardless of short surges of current, either positive or negative. A sustained current is needed to cause the voltage to vary significantly; this is equivalent to putting a steady force on the door, which will cause it to open slowly (charge up or discharge).
This is why decoupling capacitors are used around ICs, which can have very short periods of high current drain, but lower average current drain, and are fussy about their supply voltage staying steady. The capacitor holds the rail voltage steady during short bursts of high current needed by the ICs. This is the function of many of the big electrolytic capacitors on computer motherboards.
A coupling capacitor is like one of those hydraulic suppressors held between two moving points. The AC signal at one point is coupled to the other point, even though there may be DC voltage difference between the points. Very low frequencies won't "pass through" a coupling capacitor; they will cause the voltage across the capacitor to change with the signal, and the capacitor "absorbs" the signal, leaving gradual changes (low frequencies) reduced at the other end of the suppressor, so coupling capacitors have to be chosen so that the capacitance is high enough to ensure that the lowest frequencies that have to pass through the capacitor will not be attenuated (significantly) by the capacitor. DC coupled circuits (many audio amplifiers are DC coupled) have no coupling capacitors in the signal path (except usually one at the input) and as such can reproduce very low frequencies accurately.
HTH. Please re-read that referenced post and this description again if it doesn't make sense first time.
