True AC is sort of "balanced" in that it has just as much positive as negative. The positive area of the waveform is the same size as the negative area. For waveforms that are sort of symmetric across the 0V with a time offset, such as a sine or square wave, this means that it is centered along the 0V line. A DC source, on the other hand, never changes voltage.

The 0V to +10V source you have is actually a -5V to +5V square AC plus a +5V DC. The capacitor is getting rid of the DC component leaving just the AC, which happens to be the -5V to +5V AC that you are getting.

That's why the call it a DC-blocking capacitor.

The standard way of looking at this is to consider a capacitor-resistor series combination going to ground. Connect a 10v (wrt ground) supply to the capacitor and the voltage across the resistor rises to +10v, then decays. Now connect that capacitor to ground and that same resistor gets -10v across it, which then decays. Whatever is connected to the capacitor "top" terminal has to be able to sink current as well as source it.

That's what generators in simulators do - they have zero internal impedance (usually). They sink currents as well as source them.

That's why they call it a decoupling capacitor, cuz you usually just need AC (signal) at the out of a circuit, you don't need the DC part of the signal.

In my opinion an oscillator always produces an AC sine wave. There is usually no need for a DC overlapped oscillator signal. The DC supply of an oscillator produces a AC sine wave relative to GND.

Where exactly did you measure a DC sine wave, relative to what, and what do you mean by “AC removes a DC component”?

The oscillator is creating both DC and AC. The DC component is the average value of the signal. In the case of your 0-10v square wave, that is 5v. The AC compnent is the part of the signal that changes. The effect of the capacitor is to block the DC component, leaving only the AC component. The waveform is shifted vertically to be centered around 0v.