The "aggregate", "net", "combined", or "compound" transfer function of cascaded amplifier stages is sometimes quoted to be the simple product of the transfer functions of the individual stages, but this is only true if every stage implements a simple direct proportionality, like \$ V_{OUT} = -3 \times V_{IN} \$.
Your second stage introduces an offset \$ V_1 \$ which invalidates that approach.
Here you must use function composition to obtain the ouput \$ V_{OUT} \$ as a function of some input \$ V_{IN} \$:
$$ V_{OUT} = h( V_{IN} ) $$
To illustrate, I'll redraw your circuit with some extra annotations, to visually segregate the stages, and give us some variables to work with:
simulate this circuit – Schematic created using CircuitLab
The transfer functions of stages 1 and 2 are \$ f \$ and \$ g \$ respectively, so that:
$$ V_F = f(V_{IN}) $$
$$ V_{OUT} = g(V_F) $$
The two functions must be composed to obtain the overall transfer function:
$$ V_{OUT} = h(V_{IN}) = g(V_F) = g(f(V_{IN})) $$
Sometimes you see this written as \$ h = g \circ f \$, which I read as "g of f" or "g about f".
The first stage transfer function \$f\$ is very simple. I assume you are familiar with the opamp configured as a simple inverting amplifier:
$$ f(V_{IN}) = -\frac{R_9}{R_8} \times V_{IN} $$
The second stage function \$g\$ is less trivial, because of the offset \$V_A\$ due to \$V_1\$. To simplify the algebra we can employ what we know about opamps with negative feedback, which is that the opamp adjusts its output to equalise the potentials at its inverting and non-inverting inputs:
$$ V_A = V_B = 2V $$
Then construct the usual KCL and Ohm's law equations, and solve for \$ V_{OUT} \$ as a function of \$ V_F \$ and \$ V_A \$:
$$ I = \frac{V_{OUT} - V_B}{R_{11}} $$
$$ I = \frac{V_B - V_F}{R_{10}} $$
$$ V_{OUT} = \frac{R_{11}}{R_{10}}(V_A - V_F) + V_A $$
Intuitively, you can interpret this to mean that the stage amplifies the difference between \$V_A\$ and the input, with the usual gain of \$ -\frac{R_{11}}{R_{10}} \$, but also offsets the output by an additional amount \$V_A\$.
The penultimate step is to compose the functions:
$$ \begin{aligned}
h(V_{IN}) &= g(f(V_{IN})) \newline
\newline
&= \frac{R_{11}}{R_{10}}(V_A - f(V_{IN})) + V_A \newline
\newline
&= \frac{R_{11}}{R_{10}}(V_A + \frac{R_9}{R_8}V_{IN}) + V_A
\end{aligned} $$
Lastly we plug in the resistances to get:
$$ \begin{aligned}
V_{OUT} &= 2(2V + 4\cdot V_{IN}) + 2V
\newline
\newline
&= 6V + 8 \cdot V_{IN}
\end{aligned} $$