$\newcommand{\Res}{{\text{Res}}} \newcommand{\Ind}{{\text{Ind}}} \newcommand{\ds}{{\displaystyle}} \newcommand{\inv}{{^{-1}}}$By Frobenius reciprocity, if $W$ and $U$ are irreducible representations of H < G, respectively, thenthe number of times $U$ appears in $\Ind W$ is the same as the number of times $W$ appears in $\Res U$. Let $W$ be the one dimensional representation (thus obviously irreducible) of the subgroup $H$ of $G$ generated by $(1234)$ in which $(1234)\cdot w=iw$. If we want to decompose $\Ind W$ into irreducible representations of $G=S_4$ i.e how many times an irreducible representation $U$ appears in $\Ind W$, we just need to find how many times $W$ appears in $\Res U$. The trivial and alternating representations restrict to trivial representation of $C_4$ and the alternating representation of $C_4$. We know that the induced representation is $\bigoplus_{\sigma \in S_4/C_4} \sigma W $, and since $W$ is one dimensional, the induced representation has dimension $[S_4:C_4]$ which is 6. Let us begin by writing the matrix corresponding to $(1234)$ in the standard representation of $S_4$. We get this by knowing that the standard representation is just the regular representation "minus" the trivial representation. (1234) in the regular representation is just the $ \begin{pmatrix} 0 & 0 & 0 & 1 \\ 1 & 0 & 0 & 0 \\ 0 & 1 & 0 & 0 \\ 0 & 0 & 1 & 0 \\ \end{pmatrix} $ And we can write this in the basis $\begin{pmatrix} 1\\ 1 \\ 1 \\ 1 \end{pmatrix}, \begin{pmatrix} 1\\ -1 \\ 0 \\0 \end{pmatrix}, \begin{pmatrix} 0\\ 1 \\ -1 \\ 0 \end{pmatrix}, \begin{pmatrix} 0\\ 0\\ 1 \\ -1 \end{pmatrix} $ and we would get $ \begin{pmatrix} 1 & 0 & 0 & 0 \\ 0 & 0 & 0 & -1 \\ 0 & 1 & 0 & -1 \\ 0 & 0 & 1 & -1\\ \end{pmatrix} $ And thus we have that the element (1234) acts in the standard representation by the matrix $ \begin{pmatrix} 0 & 0 & -1 \\ 1 & 0 & -1 \\ 0 & 1 & -1\\ \end{pmatrix} $ which has eigenvalues $i, -i,$ and $-1$, so that the standard representation restricted to the group generated by (1234) decomposes into one of each of the irreducibles of the $C_4$ representations except for the trivial. We would find the same to be true for the standard representation tensor the alternating representation. Now since $\Res V$ and $\Res V'$ each contain a copy of $W$, we know that $\Ind W$ contains a copy of $V$ and $V'$, and since $\dim(V)=\dim(V')=3$ we need not even check the other irreducible representation of $S_4$, as $\Ind W = V \oplus V'$.
We do precisely the same for $(123)$. Let $W$ be the one dimensional representation (thus obviously irreducible) of the subgroup $H$ of $G$ generated by $(123)$ in which $(123)\cdot w=e^{2 \pi i/3} w$. We know the dimension of $\Ind W$ will be $[S_4:C_3]= 8$. In this case, We have that both the trivial and alternating representation restrict to the trivial of $C_3$.
Let us begin by writing the matrix corresponding to $(123)$ in the standard representation of $S_4$. We get this by knowing that the standard representation is just the regular representation "minus" the trivial representation. (1234) in the regular representation is just the $ \begin{pmatrix} 0 & 0 & 1 & 1 \\ 1 & 0 & 0 & 0 \\ 0 & 1 & 0 & 0 \\ 0 & 0 & 0 & 1 \\ \end{pmatrix} $
And we can write this in the basis $\begin{pmatrix} 1\\ 1 \\ 1 \\ 1 \end{pmatrix}, \begin{pmatrix} 1\\ -1 \\ 0 \\0 \end{pmatrix}, \begin{pmatrix} 0\\ 1 \\ -1 \\ 0 \end{pmatrix}, \begin{pmatrix} 0\\ 0\\ 1 \\ -1 \end{pmatrix} $ and we would get $ \begin{pmatrix} 1 & 0 & 0 & 0 \\ 0 & 0 & -1 & 1 \\ 0 & 1 & -1 & 1 \\ 0 & 0 & 0 & 1\\ \end{pmatrix} $
And thus we have that the element (1234) acts in the standard representation by the matrix $ \begin{pmatrix} 0 & -1 & 1 \\ 1 & -1 & 1 \\ 0 & 0 & 1\\ \end{pmatrix} $ which has eigenvalues $1, e^{2 \pi i/3}, e^{4 \pi i/3}$, so that the standard representation restricted to the group generated by (123) decomposes into each of the irreducibles of the $C_4$ representations We would find the same to be true for the standard representation tensor the alternating representation. Now since $\Res V$ and $\Res V'$ each contain a copy of $W$, we know that $\Ind W$ contains a copy of $V$ and $V'$, and since $\dim(V)=\dim(V')=3$ we need not even check the other irreducible representation of $S_4$, as $\Ind W = V \oplus V' \oplus Q$ just by dimensionality where $Q$ is the representation of the quotient group, it is labeled as "Another $W$" in the text by Fulton and Harris.