Mode translation descends to the affine vacuum module

For current modes, $D([x_m,y_n])=-(m+n)[x,y{]}_{m+n-1}$, whereas $[D(x_m),y_n]+[x_m,D(y_n)]=-(m+n)[x,y{]}_{m+n-1}-m((m-1)+n)B(x,y){\delta }_{m+n-1,0}K$. If the Kronecker factor is nonzero then $m+n=1$, so its central coefficient $m(m+n-1)$ is zero. Thus the two expressions agree. Brackets with $K$ also satisfy the derivation identity because $K$ is central and $D(K)=0$. Bilinearity proves that $D$ is a Lie derivation on all finite sums.

If $n\ge 1$, then $D(x_n)=-n{x}_{n-1}\in \mathfrak{g}[t]$, while $D(x_0)=0$ and $D(K)=0$. Hence $D(\mathfrak{p})\subseteq \mathfrak{p}$. Extend $D$ to the tensor algebra by the Leibniz rule. It preserves the ideal generated by $uv-vu-[u,v]$, since its derivative is a sum of the same defining relations with $u$ or $v$ replaced by $D(u)$ or $D(v)$. Therefore it descends to a derivation $D_U$ of $U(\widehat{\mathfrak{g}})$.

Define $T(u\otimes 1)=D_U(u)\otimes 1$. This respects the induced-module relation $ua\otimes 1=u\otimes a1$ for $a\in U(\mathfrak{p})$: the Leibniz rule gives the same $D_U(u)a$ term on both sides, and the remaining term $u{D}_U(a)\otimes 1$ equals $u\otimes D_U(a)1=0$ because $D(\mathfrak{p})\subseteq \mathfrak{p}$ and the vacuum character kills $\mathfrak{g}[t]$ and $D(K)=0$. Thus $T$ is well-defined. Since $D_U(1)=0$, $T\mathbf{1}=0$, and for every current mode the defining derivation action gives $[T,x_n]=D(x_n)=-n{x}_{n-1}$.