$\newcommand\Lie{\mathbf{Lie}} \newcommand\Aut{\mathbf{Aut}} \newcommand\Alg{\mathrm{Alg}} \newcommand\Grp{\mathrm{Grp}} \newcommand\Der{\mathrm{Der}}$
Let $F: \Alg \to \Grp$ be a functor and $R$ an algebra. Then $\Lie(F)(R)$ is a Lie algebra. My goal is to show that $\Lie(\Aut)(R)$ equals the Lie algebra of derivations $\Der(R).$
However, I am not sure how to define the functor $\Aut : \Alg \to \Grp.$ I can see two options:
$\Aut$ maps an algebra $A$ to its automorphism group.
We fix some algebra $R$ and define a functor $\Aut_R,$ where $\Aut_R(A)$ maps an algebra to the automorphism group of the tensor product $R \otimes A.$ From this, we construct a functor $\Aut$ which maps an algebra to the functor $\Aut_R.$ Then the goal is to prove $\Lie(\Aut(R)) = \Der(R).$
Does the first case suffice? I cannot see why the latter would be chosen, however it is used in Zhihau Chang's notes on Lie algebras of affine group schemes (see Section 4.3) and J.S. Milne's Basic Theory of Affine Group Schemes (see Ch VIII Section 2 and Ch XI).
In the second case I am struggling to understand how the morphism $\Aut_R(f)$ is defined, where $f : A \to B$ is an algebra morphism. Take $\alpha$ to be an automorphism on $R \otimes A,$ that is $\alpha \in \Aut_R(A).$ Then its image under $\Aut_R(f)$ is defined as the automorphism $\beta$ on $R \otimes B$ where $$ \beta(r \otimes 1_B) = (1_R \otimes f)\alpha(r \otimes 1_A).$$ Why does it suffice to define $\beta$ on elements $r \otimes 1_B$? Moreover, how is it surjective on $R \otimes B$ if $f$ is not necessarily surjective?