Confusion about transistor circuit analysis and voltage divider
Here's the first schematic, redrawn:



simulate this circuit &ndash; Schematic created using CircuitLab
\$Q_1\$ is a pass transistor and \$R_1\$ provides the base current needed to keep it active. If you got rid of the POT, \$Z\$, and \$Q_2\$, then the load would see \$V_\text{CC}\$, less a little voltage. But there wouldn't be any way to set the load voltage to some particular, different value.
\$Q_2\$ is added as a comparator. It compares its base voltage with its emitter voltage and adjusts its collector current so that they are different by about one diode drop. Whatever it takes. Since \$Z\$'s voltage is supposed to be fairly constant, the base voltage is what's under observation by \$Q_2\$. If the base voltage increases, then the \$V_\text{BE}\$ difference also increases and this causes \$Q_2\$ to pull much harder downward on its collector (sinking more current into \$Z\$.) But that causes the voltage drop across \$R_1\$ to increase and that causes both the base and the emitter voltages of \$Q_1\$ to decline in response. So, if \$V_\text{REG}\$ momentarily increases, then \$Q_1\$ moves back downward and therefore decreases \$V_\text{REG}\$. That's negative feedback in action.
Ideally, you'd ignore \$I_{B_1}\$ and \$I_{B_2}\$ and treat them as nil. But a good design really can't ignore them and must work out a way to make their effects small enough that they can be mostly ignored.
The worst case error will be when the POT is in its middle, halfway position as that will maximize the voltage drop due to \$I_{B_2}\$. But that may not be all that important, because you can always just turn the pot a little more in one or another direction to compensate. So that's not critical. But it's worth keeping in the back of the mind.
For a back of the envelope design and assuming that power dissipation issues can be largely set aside (it's a relatively low power circuit), you would start with the worst case load current that may be needed: \$I_{\text{LOAD}_\text{MAX}}\$. You'd assume a BJT for \$Q_1\$ that can handle that current (and then some) but defer locating it for a later time and instead assume that it can probably support \$\beta\approx 100\$ in active mode. (Again, if this is a power device that won't be true. But I'm stating at the outset this is not a power device because then you'd need a bigger envelope.)
Now, this means \$I_{B_1}\approx \frac{I_{\text{LOAD}_\text{MAX}}}{\beta\approx 100}\$. To be safe, you'll want \$I_\text{BIAS}\$ to be at least \$10\times\$ this much, though that's not cast in stone.
Now this is the point where you want to compare \$I_\text{BIAS}\approx 10\cdot\frac{I_{\text{LOAD}_\text{MAX}}}{\beta\approx 100}\$ with appropriate zener currents. If they are in a close match with each other, great. If not, and the zener current is a lot higher, then you may need to let the required zener current replace your estimate for \$I_\text{BIAS}\$ and take the hit on \$I_{B_2}\$ and the implications for \$I_\text{DIV}\$. If not, and the zener current is a lot lower, then you may want to find a different zener family because very low zener currents imply very high zener impedances and that's going to mess things up for the regulator.
A final redrawing may be helpful:



simulate this circuit
Here, I've divided this up into two sections. An upper section which moves up and down, relative to ground, and a lower section which stays solid, relative to ground. Kind of like a giraffe neck going higher or lower to reach an apple in a tree (that no giraffe is really lucky enough to eat), the upper section moves up and down based upon the voltage drop across \$R_1\$. That voltage drop is controlled by the collector current of \$Q_2\$. But that collector current is controlled by the base-emitter voltage difference of \$Q_2\$, which is measuring the output and comparing a fraction of it with the zener voltage.
And at last, we come to \$I_\text{DIV}\$. (Note that I've added \$R_2\$??) This current should also be about \$10\times\$ that needed by \$I_{B_2}\$ (worst case, again.) I've added \$R_2\$ since there's no point in losing a lot of range for the POT just because of that zener voltage. So you will want to drop about one diode drop more than the zener voltage across \$R_2\$.

There's a simple modification you can add if you want to provide a current limit capability, too.



simulate this circuit
Now you can set a current limit. Here \$Q_3\$ measures the voltage drop across IPOT. If it exceeds a certain value, \$Q_3\$ starts sinking more current via \$R_1\$ and around and behind \$Q_2\$ and into the load path. This increases the drop across \$R_1\$ and lowers the output voltage in response.