The key point is, if you disconnect the capacitor, it is still holding a charge, so it still has 9 V across it.
But when he ties BATTERY's POSITIVE terminal to CAPACITOR's NEGATIVE terminal, he reads 18.48 V! how is that possible? how capacitor acts like that enables this situation?
9 V across the battery plus 9 V across the capacitor is 18 V.
This is basic Kirchhoff's Voltage Law.
He ties BATTERY's NEGATIVE terminal to CAPACITOR's POSITIVE terminal and we read -9.14V how is this possible also?
By putting the black lead of the meter on the positive terminal of the capacitor you're choosing to make that terminal your "0 V" reference.
But there's still 9 V across the capacitor. So the negative terminal is 9 V below the positive terminal. Thus it's at -9 V.
The battery is irrelevant to this situation.
How these results can occur?
From the very basic behavior of a capacitor and Kirchhoff's Voltage Law.
How capacitor acts like this to this results to occur? (what provides this situation to occur)
The capacitor's voltage only changes if there's a current through the capacitor.
When you disconnect the capacitor from the rest of the circuit there's no current through it. Therefore the voltage across it remains at 9 V.
(If you do the experiment yourself, you'll see the voltage slowly change as you measure it due to leakage current through the meter)