I know herbschwarz has pretty much covered it, but I'll throw my rudimentary understanding in the ring too:
When we add the AC sound signal (in parallel?), this biases the AC voltage increasing its sine wave so that it should always be above zero V
That sounds generally right. In your falstad circuit, your AC source was set to go between +2.5V and -2.5V. That's 5V "peak to peak". (Note that your guitar audio source will be much less than this, maybe 1V peak to peak?)
The capacitor doesn't let current flow through it. In this case, it's sort of an "elastic membrane" for electricity: you can poke a bit of charge in one side and it will push some out the other side, but you can only pack in so much charge, before the metaphorical membrane is too stretched. At this point the "elastic" only wants to pull back to where it was, and kick out all the charge you put it, sending it back the way it came. Of course it can only do this, if it can suck some charge back in on the opposite side, to fill the "vacuum" you created there.
In this way, the charges on either side can push and pull through the "elastic membrane", without actually touching each other. If the capacitor was already balanced with 24V on the left side and 0V on the right hand side, the 24V could increase to 25V, and the 0V would only know that 1V had been "shoved out" on the right side.
This is what allows you to bias the voltage. The guitar is happily putting out it's audio signal at say +0.5V to -0.5V. The Arduino side, having been stable at 2.5V, is now feeling that pull of 0.5V up, and 0.5V down, so it will be rising to less than 3V, and falling to somewhere above 2V.
What is the affect of decreasing the resistors?
Think maybe that you're mixing the hot tap and the cold tap in the shower. If the hot tap is all the way open, the hot water is just blasting out. If you turn on the cold tap a little ways, it will make no real difference to the temperature; there is just so much hot water flowing. Now imagine you turn the hot tap back to a trickle; you can turn the cold tap just a little, and cause a big impact on the temperature of the water.
The resistors here are sort of your hot tap: the more resistance, the more the hot tap is shut. Your "cold tap" audio signal will be able to really impact the temperature. If instead you have low resistance on your "hot tap" 2.5V source, it's just POURING out, and turning the "cold tap" audio signal on and off a bit will only cause a small change in temperature.
but the amplitude of the voltage is lower with a 1u.
Whereas your "hot tap's" flow is being limited by the resistors, there is no resistor on your "cold tap" audio signal. It does however have the "rubber glove elastic membrane" capacitor in series.
If you were trying to mix a DC signal here, it wouldn't work, because the membrane would just stretch to it's limit and stop. Because you are instead putting an alternating current up against the membrane, it will quite happily push and pull. It doesn't stretch the membrane to it's limit, it just dumps a little charge in there and then pulls in back out in the other direction.
At least that's what it does when the capacitor is big enough. Say you've got a really small, taut piece of elastic. It can't be stretched as much; you can only put a tiny amount of charge it one side before it hits it's limit. If the capacitor is small, it fills up with charge so fast that even the alternating current signal can't push and pull properly. The AC signal just wants to fill the capacitor up with change for half a wave, then it wants to pull it out again, but it can't even manage this.
Maybe the capacitor is tiny, maybe the voltage is really high, or maybe the frequency is really low, pushing and pulling for a long time in each direction before alternating. Any of these could cause the capacitor to fill up and start to limit the passage of the AC signal.
I believe that this special sort of "resistance to AC signals" has it's own term, "Impedance", but this is straying into territory that I'm a bit wobbly on.
The interplay between the resistors and the impedance of your capacitor are what is defining how strong your audio signal is (peak to peak voltage), and its DC offset (the middle point of your biased wave).
Would we need to limit the output to never be above 5V?
With a large AC signal you would, however I don't see how this will be a particular issue when working with the small voltages here.
You shouldn't go above your 5V supply, but I believe that, in a pinch, the input pins will tolerate an extra 0.5V. Even if you somehow put together a circuit that put the +0.5V of the audio signal on-top of the 5V power supply, it probably wouldn't cause a meltdown. If you're unsure, double check it in falstad.
Apologies if I've gotten any part of this information dead wrong, I ain't got nun that book lernin.