Let $\mathbf{G}$ be a group functor and $R$ an algebra with inclusion map $i_R : R \to R[\epsilon],$ where $R[\epsilon]$ is the algebra of dual numbers. The adjoint action of $\mathbf{G}(R)$ on $\mathbf{Lie}(\mathbf{G})(R)$ is defined as
$$ \mathrm{Ad}(g)(x) = \mathbf{G}(i_R)(g) \cdot x \cdot \mathbf{G}(i_R)(g^{-1}),$$
for $g \in \mathbf{G}(R)$ and $x \in \mathbf{Lie}(\mathbf{G})(R).$
The adjoint action can be extended to a natural transformation $\mathrm{Ad} : \mathbf{G} \times \mathbf{Lie}(\mathbf{G}) \to \mathbf{Lie}(\mathbf{G}).$ To prove this, we need to show that for an algebra morphism $f : R \to S,$
$$ \mathrm{Ad}(\mathbf{G}(f)(g))(\mathbf{Lie}(\mathbf{G})(f)(x)) = \mathbf{Lie}(\mathbf{G})(f) (\mathrm{Ad}(g)(x)) $$
In Zhihau Chang's notes on Lie algebras of affine group schemes, Section 2.3, the proof of this is as follows, where $\tilde f : r + \epsilon r' \mapsto f(r) + \epsilon f(r')$ is the extension of $f$ to $R[\epsilon] \to S[\epsilon],$
\begin{align} \mathrm{Ad}(\mathbf{G}(f)(g))(\mathbf{Lie}(\mathbf{G})(f)(x)) &= \mathbf{G}(i_S \circ f)(g) \cdot \mathbf{G}(\tilde f)(x) \cdot \mathbf{G}(i_S \circ f)(g^{-1})\\ &= \mathbf{G}(\tilde f \circ i_R)(g) \cdot \mathbf{G}(\tilde f)(x) \cdot \mathbf{G}(\tilde f \circ i_R)(g^{-1})\\ &= \mathbf{G}(\tilde f) \left( \mathbf{G}(i_R)(g) \cdot x \cdot \mathbf{G}(i_R)(g^{-1}) \right) \\ &= \mathbf{G}(\tilde f)(\mathrm{Ad}(g)(x)) \\ &= \mathbf{Lie}(\mathbf{G})(f) (\mathrm{Ad}(g)(x)) \end{align}
I am struggling to see how the third equality holds.
My understanding is that $\cdot$ is composition in $\mathbf{Lie}(\mathbf{G})(R),$ so distributivity does not make sense.
Another reference is J.S. Milne's Basic Theory of Affine Group Schemes, Chapter XI, Section 7.
$\cdot$ here is multiplication in $\mathbf{Lie(G)}(R)$, which is the same as multiplication in $\mathbf{G}(R(\tau))$, in particular, $\mathbf{G}$ being a $k$-group functor, $\mathbf{G}(\tilde{f})$ respects that multiplication.
You have $\newcommand{\G}{\mathbf{G}} \G(\tilde{f}\circ i_R)(g)\cdot \G(\tilde{f})(x)\cdot G(\tilde{f}\circ i_R)(g^{-1}) = \G(\tilde{f})(\G(i_R)(g))\cdot \G(\tilde f)(x)\cdot \G(\tilde f)(\G(i_R)(g^{-1}) $ and so $\G(\tilde{f})$ factors out.