Electronic – Beginner Question: Get 2 Reference Voltages for Comparator using Voltage Regulators

I'm not sure I understand the configuration, especially the "not relative to ground". Does that mean the cells are in series?
In that case there's a simple solution. Connect a 1:1 resistor divider between \$V_-\$ of the lower cell and \$V_+\$ of the higher cell. \$V_+\$ of the lower cell (= \$V_-\$ of the higher cell) goes to the inverting input of the comparator; the midpoint of the divider goes to the non-inverting input. This forms a Wheatstone bridge with both cells in one branch, and the resistors in the other. If the lower cell's voltage is higher the output of the comparator will be low. Both cells will have voltages close together, so some hysteresis may be needed.

You should see immediately that the 7.09V can't be right. 7.09V on one end of the resistors and 12V on the other end can never give you 7V on the non-inverting input. Your equation for \$V_{REF}\$ seems to be wrong.

Here's how I do it. Since the current through the resistors is the same we have

\$ \begin{cases} \dfrac{V_{SAT+} - V_{UT}}{nR} = \dfrac{V_{UT} - V_{REF}}{R} \\ \\ \dfrac{V_{SAT-} - V_{LT}}{nR} = \dfrac{V_{LT} - V_{REF}}{R} \end{cases} \$

Filling in the parameters and eliminating R:

\$ \begin{cases} \dfrac{12V - 7V}{n} = 7V - V_{REF} \\ \\ \dfrac{0V - 6V}{n} = 6V - V_{REF} \end{cases} \$

From the second equation:

\$ V_{REF} = \dfrac{n + 1}{n} 6V \$

Replacing \$V_{REF}\$ in the other equation:

\$ \dfrac{12V - 7V}{n} = 7V - \dfrac{n + 1}{n} 6V \$

We find \$n =11\$, that's also the value you found. But my \$V_{REF}\$ is different:

\$ V_{REF} = \dfrac{n + 1}{n} 6V = 6.55V \$

This is OK for a theoretical calculation, but in practice you may have a problem: do you have a 6.55V source? The typical way to solve this is to get a reference voltage via a resistor divider from your 12V power supply, and then you get this circuit:

We still have 2 equations, but three unknowns, so we can choose 1. Let's pick a 30k\$\Omega\$ for R3. Then, applying KCL:

\$ \begin{cases} \dfrac{12V - V_{UT}}{R1} + \dfrac{V_{SAT+} - V_{UT}}{R3} = \dfrac{V_{UT}}{R2} \\ \\ \dfrac{12V - V_{LT}}{R1} + \dfrac{V_{SAT-} - V_{LT}}{R3} = \dfrac{V_{LT}}{R2} \end{cases} \$

Filling in our parameters:

\$ \begin{cases} \dfrac{12V - 7V}{R1} + \dfrac{12V - 7V}{30k\Omega} = \dfrac{7V}{R2} \\ \\ \dfrac{12V - 6V}{R1} + \dfrac{0V - 6V}{30k\Omega} = \dfrac{6V}{R2} \end{cases} \$

That's

\$ \begin{cases} \dfrac{5V}{R1} + \dfrac{5V}{30k\Omega} = \dfrac{7V}{R2} \\ \\ \dfrac{6V}{R1} - \dfrac{6V}{30k\Omega} = \dfrac{6V}{R2} \end{cases} \$

From which we find

\$ \begin{cases} R1 = 5k\Omega \\ R2 = 6k\Omega \end{cases} \$