Character of 2-dimensional irreducible representation of $S_4$

If $\tau: S_3\to \mathbb R$ is the character map of the standard representation then you have that the character of  the $S_4$ representation is $\tau\circ f$.

So you need to understand $f$ and $\tau$.

The standard representation of $S_3$ has a pertubation $\sigma$ act on the basis $\{e_1-e_2 , e_2-e_3\}$ by sending it to
$$\sigma(e_1- e_2) = e_{\sigma(1)}-e_{\sigma(2)},\qquad \sigma(e_2-e_3) = e_{\sigma(2)}-e_{\sigma(3)}$$
Its then elementary to calculate the matrix coefficients of this:
$$(12) \equiv \begin{pmatrix}-1&1\\0&1 \end{pmatrix},\quad (23)\equiv\begin{pmatrix}1&0\\1&-1\end{pmatrix},\quad (13) \equiv \begin{pmatrix}0&-1\\-1&0\end{pmatrix}$$
$$\mathrm{id}\equiv \begin{pmatrix}1&0\\0&1\end{pmatrix},\quad (123)\equiv \begin{pmatrix}0&-1\\1&-1\end{pmatrix},\quad (132)\equiv\begin{pmatrix}-1&1\\-1&0\end{pmatrix}$$
Ans an example calculation:
$(12) (e_1-e_2) = e_2-e_1=-(e_1-e_2), \quad (12)(e_2-e_3) = e_1-e_3  =(e_1 - e_2) + (e_2-e_3)$
which then gives the above matrix coefficients. 

Then one gets the following trace-map on $S_3$:
$$\tau(12)= \tau(23)=\tau(13) = 0, \qquad \tau(123)=\tau(132) = -1,\qquad \tau(\mathrm{id})=2$$

To calculate the induced trace-map on $S_4$ we need to find the pre-images of the above elements under $f$. There are $24$ elements of $S_4$, so this is probably a bit akward. Its better to note that the character only depends on the conjugacy class of a permutation, there are only $5$ different conjugacy classes so thats nice.

Lets start with the transpositions in $S_4$ these are:
$$(12), \ (23), \ (34),\ (13), \ (14), \ (24)$$
and they give one conjugacy class. Note that $f(12)=(12)$ so that $\tau(f(12))=0$. Hence the character is $0$ on all transpositions.

Next we look at $\mathrm{id}$, this is also a conjugacy class and $\tau(f(\mathrm{id}))=2$.

Now the cycles of size $3$ are another conjugacy class, they are:
$$(123), (132), (234), (243), (341), (314), (412), (421)$$
Now the image of $f(123) = f((23)(12)) = (13)(12) =(123)$ which gets sent under $\tau$ to $-1$. So all the above get sent to $-1$ under the character.

Now there are $6$ cycles of size $4$:
$$(1234), (1243), (1324), (1342),(1423), (1432)$$
we have $f(1234) = f((34)(23)(12)) = (12)(13)(12)= (13)$, which gets sent to $0$ under $\tau$ so the character is also $0$ here.

Next there are $4$ elements:
$$(12)(34), (13)(24), (14)(23)$$
we have eg $f((12)(34)) = (12)(12)=\mathrm{id}$. So these get sent to $2$ under the character.

Finally we have $\mathrm{id}$, which gets sent to $2$ under the character.