Let $X$ be a scheme over an algebraically closed field $k$ and let $\operatorname{Aut}(X)$ denote the functor sending a $k$-scheme $T$ to the group $\operatorname{Aut}_T(X \times_k T)$ of automorphisms of $X \times_k T$ over $T$.

My goal is to have a better grasp of why $\operatorname{Lie}(\operatorname{Aut}(X))= H^0(X, \mathcal{T} X)$
and therefore I am trying to work through an example where I know both the group $\operatorname{Aut}(X)$ and $\operatorname{Lie}(\operatorname{Aut}(X))$. For example, I know that intuitively $H^1(X, \mathcal{T}X)$ gives all possible ways in which we may glue together the trivializing cover of a deformation of $X$ over the dual numbers. I want to figure out an analagous statement for $H^0(X, \mathcal{T}X)$.

Let $X = \mathbb{P}_k^1$ so that $\operatorname{Aut}(X)= PGL(2,k)$. Now I try to recover the fact that $\operatorname{Aut}(X)= PGL(2,k)$.

The global sections of $X$ are locally of the form $a_0 \partial_z + a_1 z \partial_z+ a_2 z^2 \partial_z$ where $z=v/u$ is a choice of homogeneous coordinates on $X$. This should somehow be related to the morphisms $$ z \mapsto \frac{az+b}{cz + d}.$$

Is it possible to go from this description of global sections to the group $\operatorname{Aut}(X)$ as written above?

Solving for the integral curve I end up with the equation $z'(t) = a_0 + a_1 z(t) + a_2z^2(t)$. If $a_0=0$, this would be a Bernoulli differential equation and I could solve it..