Let $ H=S_2 \subset G=S_3 $. Then use Frobenius reciprocity to decompose $ \operatorname{Ind}_H^G(\operatorname{sgn}_H) $ into irreducibles.
$ G=S_3 $ has $ 3 $ irreps $ 1_G, \operatorname{sgn}_G$ and a two-dimensional irrep $ \pi $. The character table is :
\begin{array}{c|c|c|c} \# & 1 & 3 & 2 \\ & e & (12) & (123)\\ \hline \rho_1 = 1_G & 1 & 1 & 1 \\ \rho_2 = \operatorname{sgn}_G & 1 & -1 & 1 \\ \rho_3 = \pi & 2 & 0 & -1 \end{array}
I believe that the irreps of $S_2$ are $\{e, (12)\}$
For $H \subseteq G$, if $V$ is an irreducible representation of $H$ and $W$ is an irreducible representation of $G$, then Frobenius reciprocity says that $$ \mathrm{mult}_W(\mathrm{Ind}_H^G V) = \mathrm{mult}_V(\mathrm{Res}^G_HW). $$ So we only need to determine the multiplicity of $\mathrm{sgn}_{\mathfrak{S}_2}$ in the restrictions of the irreducible representations of $\mathfrak{S}_3$
Recall that $\dim(\mathrm{Ind}_H^G) = [G:H] =3$. Considering that two of the three representations of $\mathfrak{S}_3$ are $1$-dimensional, there is not a lot of choice.
Certainly $\mathrm{sgn}_{\mathfrak{S}_2}$ does not appear in the trivial representation of $\mathfrak{S}_3$ (since the non-trivial element of $\mathfrak{S}_2$ is odd but it acts trivially the trivial representation).
The sign representation of $\mathfrak{S}_3$ restricts to the sign representation of $\mathfrak{S}_2$, so it provides $1$ copy of $\mathrm{S}_{\mathfrak{S}_2}$.
We can do the calculation and see that the standard representation of $\mathfrak{S}_3$ splits into two copies of the sign representation of $\mathfrak{S}_2$. But even without doing the calculation, we have to reach get $3$ copies of $\mathrm{sgn}_{\mathfrak{S}_2}$ among the irreducibles of $\mathfrak{S}_3$: we have one from the sign representation so there are two left.
If $V = \mathrm{sgn}_{\mathfrak{S}_2}$, we conclude $$ \mathrm{mult}_{V}(\mathrm{Res}^{\mathfrak{S}_3}_{\mathfrak{S}_2} \ \mathrm{triv}_{\mathfrak{S}_3}) = 0\\ \mathrm{mult}_{V}(\mathrm{Res}^{\mathfrak{S}_3}_{\mathfrak{S}_2} \ \mathrm{sgn}_{\mathfrak{S}_3}) = 1\\ \mathrm{mult}_{V}(\mathrm{Res}^{\mathfrak{S}_3}_{\mathfrak{S}_2} \mathrm{standard}_{\mathfrak{S}_3}) = 2\\ $$ so by Frobenius reciprocity $$ \mathrm{Ind}^{\mathfrak{S}_3}_{\mathfrak{S}_2} \ \mathrm{sgn}_{\mathfrak{S}_2} = \mathrm{sgn}_{\mathfrak{S}_3} \oplus \mathrm{standard}_{\mathfrak{S}_3}. $$