The reason that no supply is shown connected to U3.2, is that it shares the power supply of U3.1, which is part of the same integrated circuit. The TL084 contains 4 op-amps, all sharing the same supply on pins 4 and 11, as shown here (from elprocus.com):
Therefore, it is powered, even though the schematic seems to show that it isn't.
Op-amp U3.2 has its non-inverting input connected to some non-zero potential \$V_{S\_REF}\$, which causes its output to be offset (centered about) some non-zero value also.
As I explain in this answer, for the first (U3.2) stage, the relationship between input and output is:
$$ V_{OUT} = V_{S\_REF} \left( 1 + \frac{R_{10}}{R_8} \right) - V_{DAC\_X}\frac{R_{10}}{R_8} $$
Plugging in known values gets you:
$$
\begin{aligned}
V_{OUT} &= V_{S\_REF} \left( 1 + \frac{12k\Omega}{3k\Omega} \right) - V_{DAC\_X}\frac{12k\Omega}{3k\Omega} \\ \\
&= 5 V_{S\_REF} - 4 V_{DAC\_X} \\ \\
\end{aligned}
$$
As you can see, the inverting input \$V_{DAC\_X}\$ is subject to gain −4. The non-inverting input \$V_{S\_REF}\$ is a constant potential, subject to gain +5, and the goal now is to determine what fixed potential there would produce the required offset. First rearrange the above equation to have \$V_{S\_REF}\$ the subject:
$$
\begin{aligned}
V_{OUT} &= 5 V_{S\_REF} - 4 V_{DAC\_X} \\ \\
V_{S\_REF} &= \frac{V_{OUT} + 4 V_{DAC\_X}}{5}
\end{aligned}
$$
Then plug in some known values to reveal \$V_{S\_REF}\$. We know, for instance, that when the input is \$V_{DAC\_X}=+2V\$, the output should be \$V_{OUT}=-4V\$:
$$
\begin{aligned}
V_{S\_REF} &= \frac{(-4V) + 4 (+2V)}{5} \\ \\
&= +0.8V
\end{aligned}
$$
That's why you require R5 and R6 to produce \$V_{S\_REF}=+0.8V\$.
The second stage (U3.1) is a simple inverting amplifier, with gain \$-\frac{R_{14}}{R_{11}}=-1\$ so that as the prior stage output goes from −4V to +4V, the second output does the exact opposite, going from +4V to −4V.
To have the system produce the same output voltage range for inputs between 0V and +3.3V, you can use the same algebra I used above. One trick you can use to get started is to recognise that input \$V_{DAC\_X}\$ is subject to gain \$-\frac{R_{10}}{R_8}\$. If you can find that gain, then you can choose R10 and R8 to suit. Gain \$A\$ is the ratio output swing to corresponding input swing:
$$
\begin{aligned}
A &= \frac{(-4V) - (+4V)}{(0V) - (+3.3V)} \\ \\
&= \frac{-8}{3.3} \\ \\
\end{aligned}
$$
Use that value to obtain R10 and R8. If I leave R10 unchanged:
$$
\begin{aligned}
-\frac{R_{10}}{R_8} &= -\frac{8}{3.3} \\ \\
R_8 &= 3.3\frac{R_{10}}{8} \\ \\
&= 3.3\frac{12k\Omega}{8} \\ \\
&= 4.95k\Omega \\ \\
\end{aligned}
$$
From there you can use the same procedure as above to determine the required \$V_{S\_REF}\$.
