The adjoint action preserves the associated graded of a two-sided ideal

$I\cap F^{n}U(\mathfrak{g})$ defines an increasing filtration of $I$ with $I\cap F^{-1}U(\mathfrak{g})=0$, so $\mathrm{gr}I$ is a graded subspace of $\mathrm{gr}U(\mathfrak{g})$ by construction. It is an ideal: if $a\in I\cap F^{m}U(\mathfrak{g})$ and $b\in F^{n}U(\mathfrak{g})$, then $ab\in I\cap F^{m+n}U(\mathfrak{g})$ by [F4], and $ba\in I\cap F^{m+n}U(\mathfrak{g})$ likewise; passing to symbols with [F1] exhibits every product of a symbol of $I$ with a symbol of $U(\mathfrak{g})$ as a symbol of an element of $I$. Since $\mathrm{gr}U(\mathfrak{g})=S(\mathfrak{g})$ is commutative by [F3], one-sided closure suffices and $\mathrm{gr}I$ is a graded ideal.

The commutator map ${\mathrm{ad}}_x\text{\hspace{0.17em}⁣}:U(\mathfrak{g})\to U(\mathfrak{g})$, $u\mapsto xu-ux$, is a derivation of $U(\mathfrak{g})$: ${\mathrm{ad}}_x(uv)=(xu-ux)v+u(xv-vx)={\mathrm{ad}}_x(u)v+u{\mathrm{ad}}_x(v)$. For a word $a_1\cdots a_m$ with $a_i\in \mathfrak{g}$, the derivation rule gives $[x,a_1\cdots a_m]={\sum }_{i=1}^m{a}_1\cdots a_{i-1}[x,a_i]a_{i+1}\cdots a_m$, and each $[x,a_i]$ lies in $\mathfrak{g}$; hence every term still has PBW degree at most $m$, so ${\mathrm{ad}}_x(F^{n}U(\mathfrak{g}))\subseteq F^{n}U(\mathfrak{g})$. It maps $I$ into itself because $I$ is two-sided.

By step 1.2, ${\mathrm{ad}}_x$ preserves each $I\cap F^{n}U(\mathfrak{g})$, so it induces a graded linear map $\mathrm{gr}({\mathrm{ad}}_x)$ of $\mathrm{gr}U(\mathfrak{g})$ that sends $\mathrm{gr}I$ into itself: the induced map is ${\sigma }_n(u)\mapsto {\sigma }_n({\mathrm{ad}}_{x}u)$, well defined because ${\mathrm{ad}}_x(F^{n-1})\subseteq F^{n-1}$. The derivation identity of step 1.2 passes to symbols via [F1], so $\mathrm{gr}({\mathrm{ad}}_x)$ is a derivation of $\mathrm{gr}U(\mathfrak{g})=S(\mathfrak{g})$.

On degree one, $\mathrm{gr}({\mathrm{ad}}_x)(y)={\mathrm{ad}}_x(y)=[x,y]$ for $y\in \mathfrak{g}$, by [F2] and [F4]; that is, the induced derivation restricts on $\mathfrak{g}$ to the given linear map ${\mathrm{ad}}_x$.

A derivation of $S(\mathfrak{g})$ is determined by its values on $\mathfrak{g}$: on a monomial $y_1\cdots y_n$ the Leibniz rule forces ${\sum }_i{y}_1\cdots {\mathrm{ad}}_x(y_i)\cdots y_n$, and a monomial basis of $S(\mathfrak{g})$ extends these values linearly. Hence the derivation $\mathrm{gr}({\mathrm{ad}}_x)$ of step 2.1, whose degree-one restriction is the given map ${\mathrm{ad}}_x$ by step 3.1, equals $D_x$. Therefore $D_x(\mathrm{gr}I)=\mathrm{gr}({\mathrm{ad}}_x)(\mathrm{gr}I)\subseteq \mathrm{gr}I$ and ${\sigma }_n({\mathrm{ad}}_{x}u)=D_x({\sigma }_n(u))$ for every $n\ge 0$ and $u\in F^{n}U(\mathfrak{g})$, which is the assertion.