The input is high-impedance and as such hardly draws any current. But let's, for sake of argument, pretend there flows a (rather large) current of 1\$\mu\$A. This current will flow through the 10k\$\Omega\$ pull-up resistor causing a 10mV (1\$\mu\$A \$\times\$ 10k\$\Omega\$) voltage drop across it. So in this case the voltage on the input pin will be \$V_{CC}\$ - 10mV, probably 5V - 10mV = 4.99V. That will be still recognized as a high level, so no problems here.
The 10k\$\Omega\$ is a typical value for pull-up resistors for this reason: even if there's a small leakage current the voltage drop is negligible. Don't be tempted to increase it to 1M\$\Omega\$, though it will decrease the current when the switch is closed. At 1\$\mu\$A leakage current the voltage drop will be 1\$\mu\$A \$\times\$ 1M\$\Omega\$ = 1V, and then the 5V will drop to 4V. For a 5V supply this will still be OK, but for a 3.3V supply the resulting 2.3V may be too low to be always seen as a high level.
For the pull-down the story is about the same. There doesn't flow any current in the input; you can't say that it would be connected to ground (in which case closing the switch would indeed cause a short-circuit). As such the input takes the voltage you apply to it. If the switch is closed this is \$V_{CC}\$. If the switch is open it's ground (through the pull-down resistor). If there's no current flowing (ideal world) then there's no voltage drop across the resistor either, and the input will be at \$GND\$ level. In a real world situation it may be a few mV.
First: Yes, your understanding is essentially correct, other than the issue being voltage and not charge.
Here is my analogy:
Consider a door to a house, with really smooth hinges, and no bolt or latch. The door is so light and so well-hinged that the slightest breeze would cause it to flap open and closed.
Now add a light door-spring to the door. The spring keeps the door shut, but not terribly firmly: A gentle push will open it, and letting it go will cause the door to close again.
A so-called "floating input" is like that door - the slightest perturbations in electromagnetic field, like the breeze above, will cause the input to randomly toggle between open and shut (low and high).
Add the pull-up resistor (if you want the default to be "high") or pull-down resistor (if you want it to be "low"), and your spring is in place.
Now, an external voltage applied, like the gentle push, can overcome the "keep the door shut" tendency of the spring / pull-x resistor - and once the push is removed, the input returns to the desired default value.
A low value resistor in such use is like a really stiff spring - it needs a much firmer push to open, but open it will. It will also slam shut faster when the push is removed.
Best Answer
The high output is specified as a minimum of 2.4V. This is enough if the ATmega is running off a supply of 4V or less, otherwise a pullup may be needed to pull it up to 0.6VCC.