Following an email from Efimov, I can now complete this direct check.
First let's give the maps names. The first SES is
$$A \xrightarrow{f_1} B \xrightarrow{f_2} C$$
while the second is
$$C \xrightarrow{f_3} D \xrightarrow{f_4} E.$$
The first SES corresponds to the accessible functor $f_2^{RR}f_1$ and the second to $f_4^{RR}f_3$ (where as in the question $R$ superscript means taking right adjoint).

The composition of these two correspondences is the subcategory in
$B \amalg_{C} D$ generated by $A$ and $D$.
To find the accessible functor this composed correspondence corresponds to, we just need to compute
$$\operatorname{Hom}(i_Df_4^Re,i_Bf_1a)$$
for $e \in E$ and $a \in A$ where $i_B$, $i_D$ are the inclusion functors into the pushout category.

So the goal is to compute the composition $f_4^{RR}i_D^Ri_Bf_1$.

Now, we know that $B \coprod_C D \cong B \times_C D$ with the right adjoint functors (so $f_2^{RR}$ and $f_3^R$).
We can compute that
$$i_B : B \to B \coprod_C D \cong B \times_C D$$
is defined by
$b \mapsto (b,f_3f_2^{RR}b,-)$ where $-$ stands for the natural homotopy between $f_2^{RR}b$ and $f_3^Rf_3f_2^{RR}b$.

We also know that
$$i_D^R : B \times_C D \to D$$ is simply the second projection. Hence, the functor we need to compute is $f_4^{RR}f_3f_2^{RR}f_1$ as desired.

Note (as an aside) that $i_D^L$ the left adjoint is induced by $f_2 \amalg id$.

Addendum:
The diagram in $\mathrm{Pr}^{LL}_\text{St}$
$\require{AMScd}$
\begin{CD}
A @>{f_1}>> B @>{f_2}>> C \\
@V{\cong}VV @V{i_B}VV @VV{f_3}V \\
A @>>> B \amalg_C D @>{i_D^R}>> D \\
\end{CD}

has enough data in $\mathrm{Pr}^L_\text{St}$ to make the pushout an absolute $\mathrm{Stab}$-enriched colimit. (Just like short exact sequence in $\mathrm{Pr}^{LL}_\text{St}$ which is a split-exact sequence in $\mathrm{Pr}^L_\text{St}$.)

Hence it is also a pushout in presentable categories with accessible functors. This is enough to pushout the $f_2^R \vdash f_2^{RR}$ adjunction along $f_3$ to conclude that
\begin{CD}
B @>{f_2^{RR}}>> C \\
@V{i_B}VV @VV{f_3}V \\
B \amalg_C D @>{i_D^{RR}}>> D \\
\end{CD}
from where we can finish as above.