The transposition class sum acts on a complex Specht module by total content

Conjugation by any permutation sends $(ab)$ to the transposition of its two images, hence permutes the summands of $T_m$. Thus $T_m$ is central, so its action on the nonzero finite-dimensional irreducible Specht module is an intertwiner and equals $z\mathrm{id}$ for some $z\in \mathbb{C}$ by [F2]. Choose the row-filled tableau $t$ of shape $\nu$, and use its nonzero polytabloid $e_t$. Taking the coefficient of $\{t\}$ in $T_m{e}_t=z{e}_t$ recovers $z$, since that coefficient in $e_t$ is $1$ by [F1].

A summand indexed by a transposition $\tau =(ab)$ and $\gamma \in C_t$ contributes to this coefficient precisely when $\tau \gamma \in R_t$. Put $r=\tau \gamma$, so $\tau =r{\gamma }^{-1}$. If $x\mathrm{\notin }\{a,b\}$, then $r{\gamma }^{-1}(x)=x$. The entry ${\gamma }^{-1}(x)$ has the column of $x$, and since $r$ preserves each row it also has the row of $x$. Their row-column intersection consists of $x$ alone, so ${\gamma }^{-1}(x)=x$ and $r(x)=x$. Thus both $r$ and $\gamma$ fix the complement of $\{a,b\}$; each is either the identity or $\tau$. Their product is $\tau$ exactly when one is $\tau$ and the other is the identity. Therefore the contributing pairs are exactly: $\gamma =1$, $\tau \in R_t$, of sign $+1$; and $\gamma =\tau$, $\tau \in C_t$, of sign $-1$. These cases cannot overlap, since two distinct entries cannot share both row and column.

There are ${\sum }_i\left(\genfrac{}{}{0px}{}{{\nu }_i}{2}\right)$ transpositions within rows and ${\sum }_j\left(\genfrac{}{}{0px}{}{{\nu }_j^{\mathrm{\prime }}}{2}\right)$ within columns. By steps 1.1 and 2.1 their difference is $z$. Summing $c-1$ within every row gives the first count, and summing $r-1$ within every column gives the second; hence their difference is ${\sum }_{(r,c)\in [\nu ]}(c-r)$. Also $n(\nu )={\sum }_{(r,c)\in [\nu ]}(r-1)$ and $n({\nu }^{\mathrm{\prime }})={\sum }_{(r,c)\in [\nu ]}(c-1)$, proving every formula for $z_{\nu }$. This includes $m=0,1$: there are no row or column pairs and the transposition sum is zero.

For $m\ge 2$, all $\left(\genfrac{}{}{0px}{}{m}{2}\right)$ transpositions are conjugate, so their representing matrices are similar and have character value ${\chi }^{\nu }((12))$. Taking traces of the scalar action in step 3.1 gives $\left(\genfrac{}{}{0px}{}{m}{2}\right){\chi }^{\nu }((12))=d_{\nu }{z}_{\nu }$ by [F3]. Dividing by the nonzero integer $\left(\genfrac{}{}{0px}{}{m}{2}\right)$ proves the stated character formula. The construction and coefficient count are finite and use no choice of an arbitrary family or seminormal-basis input.