My question is, how might I construct the tangent bundle on a smooth scheme?
It is clear how to define the tangent space at a point: the Zariski tangent space.
It is also clear what we should do in the setting of manifolds: we assemble the tangent spaces at all the stalks and put a topology on it. This, along with the evident projection map, give a vector bundle.
But the situation for schemes is perhaps more difficult.
The other question I have is about why this construction is not more well known or used in algebraic geometry.
Best Answer
Question: "My question is, how might I construct the tangent bundle on a smooth scheme? It is clear how to define the tangent space at a point: the Zariski tangent space."
Answer: If $k$ is a field and $k \rightarrow A$ is a $k$-algebra, let $I\subseteq A\otimes_k A$ be the kernel of the multiplication map $m:A\otimes_k A \rightarrow A$ defined by $m(a\otimes b):=ab$. It follows $\Omega:=\Omega^1_{A/k}:=I/I^2$ is the module of Kahler differentials of $A$. If $\mathfrak{m} \subseteq A$ is a maximal ideal with residue field $\kappa(\mathfrak{m})\cong k$ it follows
$$\Omega^1 \otimes_A \kappa(\mathfrak{m}) \cong \mathfrak{m}/\mathfrak{m}^2$$
is the cotangent space at the ideal $\mathfrak{m}$. Hence the cotangent module has at any $k$-rational point $\mathfrak{m}$ the cotangent space as fiber. For this reason we call $\Omega$ the "cotangent module" of $A$. When $\Omega$ is a finite rank locally free $A$-module it follows its associated vector bundle $X:=\mathbb{V}((\Omega^1)^*):=Spec(Sym_A^*((\Omega^1)^*)$ is the cotangent bundle on $S:=Spec(A)$. There is a canonical map $\pi: X \rightarrow S$ which is locally trivial in the following sense: There is an open affine cover $U_i$ of $S$ with $\pi^{-1}(U_i) \cong \mathbb{A}^r \times U_i$ where $r:=rk(\Omega^1)$. These definitions make sense for any scheme $X/S$ using the diagonal $\Delta: X \rightarrow X\times_S X$.
The dual $(\Omega^1)^*\cong Der_k(A)$ is the module of derivations. Its associated vector bundle
$$\Theta_{S/k}:=\mathbb{V}(Der_k(A)^*)$$
is the tangent bundle of $S$. When $Der_k(A)$ is locally free there is for any $k$-rational $\mathfrak{m}$ point an isomorphism
$$Der_k(A)\otimes_A \kappa(\mathfrak{m}) \cong Hom_{\kappa(\mathfrak{m})}(\mathfrak{m}/\mathfrak{m}^2, \kappa(\mathfrak{m})),$$
hence the fiber of the module of derivations is the Zariski tangent space.