Why does the voltage of a zener diode match the voltage at Vout?

Zener diode passes a lot of current (has low resistance) if the voltage is above Vz. So, connected in a circuit like in the question:

1. If the voltage at the output was lower than Vz, the diode would not conduct and (if there was o other load, the voltage would rise.
2. If the voltage at the output was higher than Vz, the diode would have low resistance and short out the output, which means that the voltage would drop (because of the resistor in series).

So, the circuit reaches an equilibrium of Vout=Vz. If the voltage tries to go lower (a load is connected), the diode conducts less and the voltage rises back up to Vz. If the voltage tries to go higher (aload was disconnected) then the diode would conduct more and drop the voltage down to Vz.

Does that have to do with Kirchhoff's Voltage Law?

Yes, by KVL, \$V_{out} = V_z\$.

How does the zener diode, if it drops 5V to ground, cause Vout to also be 5V?

Another way to 'see' this is the load (not shown) is in parallel with the zener diode and, as you know, and by KVL, parallel connected circuit elements have identical voltages.

The circuit with load:

^{simulate this circuit – Schematic created using CircuitLab}

See that the zener diode and load are parallel connected and thus, the voltage across the load and voltage across the zener diode are identical:

$$V_{out} = V_z $$

Rhetorical question: What's the current through the zener?

This might help:

As you can see, once the voltage hits the zener voltage (-17.1V in this case), the zener starts conducting a lot of current, really fast.

Now what's the voltage across the resistor?

We know from Ohm's law that the voltage across the resistor will be \$E = IR\$. So as the zener starts conducting a lot of current, the current through the resistor will go up also, and by Ohm's law, so will the voltage across the resistor.

When the voltage across the resistor increases, the voltage across the zener must decrease (that's Kirchoff's voltage law). If it were to decrease too much, the zener would stop conducting, and then there would be no current, so the resistor voltage would fall to zero, and the zener would again see \$V_{in}\$ and conduct.

Thus an equilibrium is reached: the zener conducts just enough current to make the voltage drop across the resistor equal to \$V_{in} - V_z\$.

A zener regulator such as this has two problems. Firstly, quite a lot of current can flow in the resistor and the zener, even when no load is connected. This is inefficient.

Secondly, as the load draws more current, the zener has to conduct less current, because the current through the resistor (which determines the output voltage) is the sum of the zener current and the load current. At some point, the load draws enough current on its own such that the zener doesn't have to conduct any. If the load current is further increased, the voltage drop across the resistor will be too much, and since the zener can't source current (only sink it), \$V_{out}\$ becomes less than \$V_z\$.