How to add specific voltage offset to the "balanced phase splitter" op amp circuit?
Overview
Okay. You've complicated things a bit, though I suppose I should have been able to infer that both outputs were to be similarly adjusted if I'd given it more thought. (A reason for you to also consider writing more thoroughly when you write, too.)
I first considered whether it may be possible to retain something of your existing approach. The answer is yes, but as with everything it comes with more questions. I'm going to ignore those questions in the interest of saving both our times. (You can always write up another question and provide better information, if I miss the mark.)
So I'm going to suggest a simple resistor arrangement that will shift the input source's voltage range. (I'm assuming it can drive the load. One of those questions I already asked you and also for which I don't think I received an answer.) This can be done with just three resistors. The problem is attenuation. So this the suggests adding some voltage gain to compensate for the loss.
That done, the next thing is your second stage. It needs to account for the shifted output of the first stage. The result is that both opamps will be operating their outputs almost entirely towards the upper half of their supply range. Which is probably okay. But again, no more questions. Just move forward.
Stage 1 Overview
The basic topology for the first stage is:



simulate this circuit &ndash; Schematic created using CircuitLab
The resistor network supplying the (+) input of the opamp is shown above. Also included are resistors needed for setting the voltage gain, too. We don't yet know the approximate voltage gain -- not until we have some idea about the three resistors feeding the (+) input -- so that's going to be the first step.
This isn't the kind of design that goes from step A to step B to step C, though. There's some back-and-forth thinking required. A fact of life that is ever in the minds of engineers throughout their lives. You almost never get to just walk through a linear path of steps, in practice.
One thing I do know that will be a requirement as I move forward is that the (+) input cannot go below zero volts. That's because the above design will just apply its gain to that and we already know you want only positive or zero output voltage of this stage. This helps bound the issue. I also plan to use the E24 series of values. Another bound that I hope won't bite me, later.
Stage 1 Design
Just off the cuff, I'm going to use \$R_5=47\:\text{k}\Omega\$. I expect the Thevenin resistance to be about a third of that and that's fine for use with a general opamp. (More questions I have that I won't ask.) Given that, I find that \$R_4=56\:\text{k}\Omega\$ and \$R_3=37.925\:\text{k}\Omega\$ would achieve an exact zero volts. (Okay. Almost. It's really a minimum of \$+11.47\:\mu\text{V}\$. Close enough.)
But there's a little more trouble still. Resistors have tolerances. (Like pretty much everything.) E24 and E48 series resistors do come in 2%, for example. And 1% resistors are readily available with E96 series values. So let's see what 2% does if \$R_3=36\:\text{k}\Omega\$, \$R_4=56\:\text{k}\Omega\$, and \$R_5=47\:\text{k}\Omega\$. In this case, the (+) input of the opamp would observe, nominally (assuming perfectly exact resistor values), \$316.1\:\text{mV}\$ to \$4.132\:\text{V}\$. But given 2% variation in the resistors and the associated uncertainty, the minimum could be as little as \$75.5\:\text{mV}\$ and the maximum as much as \$4.332\:\text{V}\$.
The actual range of the three-resistor divider's output node, taking into account 2% resistor variation, is:
$$\begin{align*}
  \begin{array}{r}
  {\text{minimum}:}\vphantom{75.498376\:\text{mV}}\\\\
  {\text{nominal}:}\vphantom{75.498376\:\text{mV}}\\\\
  {\text{maximum}:}\vphantom{75.498376\:\text{mV}}
  \end{array}
&amp;&amp;
\overbrace{
 \begin{array}{r}
      75.498376\:\text{mV}\\\\
      316.08833\:\text{mV}\\\\
      558.30068\:\text{mV}
 \end{array}
  }^{\text{Input}:\quad -6.0\:\text{V}}
&amp;
  \begin{array}{c}
  &amp;\quad\vphantom{558.30068\:\text{mV}}\\\\
  &amp;\quad\vphantom{558.30068\:\text{mV}}\\\\
  &amp;\quad\vphantom{558.30068\:\text{mV}}
  \end{array}
&amp;
\overbrace{
 \begin{array}{l}
      3.9277850\:\text{V}\\\\
      4.1318612\:\text{V}\\\\
      4.3322039\:\text{V}
 \end{array}
  }^{\text{Input}:\quad +6.0\:\text{V}}
\end{align*}$$

After applying gain to this, it may leave something to be desired in
the end. On the other hand, who am I to know? You've not specified how
tightly all this needs to be controlled.

I'm going to stick with the above three resistor values for now. Good enough to get the point across.
Given that you want about a \$12\:\text{V}\$ span from bottom to top (that's one way of interpreting your specs -- there are others, such as requiring the top to be exactly \$12\:\text{V}\$ but then letting up on the span from bottom to top, etc) and now knowing that I nominally have \$4.132\:\text{V}-316.1\:\text{mV}=3.816\:\text{V}\$ the voltage gain then needs to be \$\frac{12\:\text{V}}{3.816\:\text{V}}\approx 3.145\$. I also want about the same Thevenin impedance, so I'm selecting \$R_1=47\:\text{k}\Omega\$ and \$R_2=22\:\text{k}\Omega\$ which gets me very close.
So, we now have:



simulate this circuit
Given the values I've used above, I'd expect the output to vary from \$\approx 1\:\text{V}\$ to \$\approx 13\:\text{V}\$. But that's nominally. With 2% variation in the resistors? Both the offset and the gain can be more different than you'd like them to be.
You can repair both of these problems by using two added potentiometers, one to adjust for offset and one to adjust the voltage gain. In fact, I almost think it would not be an option, but a necessity if you really want what it appears you want.
Stage 2
Assuming you got this far, you must by now know that your second stage will also require some adjustment. You've already learned about subtracting current from a node to move things around. So I'll leave this one to you.
Signing off...
Note Regarding the 3R Network
If interested, I just cobbled up a quick VB.NET program that will allow others to easily answer 3R questions. I'm providing both full source (no need to trust the .EXE) and a pre-compiled .EXE for Windows for those not interested in self-compiling the source. It's in a ZIP file just called R3. The display looks like this:

Perhaps it may help.
It's just a one-off. I was curious. So I did it today. I could have improved a number of things and the only time I spent on UI was in getting the needed pieces there, little else. You can use SI units with your numbers. So 47k is just fine. (You don't need to use 47E3 or 47000.) That said, I'm not offering to re-write it in JavaScript or Blazor and WebAssembly. Feel free if you want. Most of the necessary stuff is there.