= Solution
Put $d_i=\bar X_{n-1,i}-\bar X_n=(\bar X_n-X_i)/(n-1)$. Taylor's theorem gives
$$
T_{n-1,i}-T_n=g'(\bar X_n)d_i+R_{n,i},
\qquad |R_{n,i}|\leq M d_i^2.
$$
Since $\sum_i d_i=0$, centering the leave-one-out values and expanding the square gives
$$
v_{\rm JACK}=\frac{n-1}{n}(A_n+B_n+2C_n),
$$
with the stated $A_n,B_n$ and the cross term satisfying $|C_n|\leq\sqrt{A_nB_n}$ by Cauchy--Schwarz.
Now
$$
A_n=\frac{g'(\bar X_n)^2}{(n-1)^2}
\sum_i(X_i-\bar X_n)^2
\sim\frac{g'(\mu)^2\sigma^2}{n}.
$$
The fourth-moment assumption gives
$$
B_n\leq\frac{M^2}{(n-1)^4}\sum_i(X_i-\bar X_n)^4=O(n^{-3}),
$$
and hence $C_n=O(n^{-2})$. Since $\sigma_n^2=g'(\mu)^2\sigma^2/n$, it follows that
$$
\boxed{\frac{v_{\rm JACK}}{\sigma_n^2}\to1.}
$$
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