Edit: It seems my language was not correct. By "orientable", I mean there are no non-zero sections of the tangent bundle, i.e. the sphere fails the Hairy Ball Theorem, while the circle does not.
Let $TS^1$ and $TS^2$ be the tangent bundles to the circle $S^1 \subset \mathbb R^2$ and the sphere $S^2 \subset \mathbb R^3$ respectively.
It's a "well-known" fact that $TS^1$ is an orientable, real two-manifold, while $TS^2$ is an unorientable, real four-manifold.
My question: Can I establish this from their equations?
Taking $xy$-coordinates on the plane, and $uv$-coordinates for the tangent space at the point $(x,y) \in S^1 \subset \mathbb R^2_{(x,y)}$, I can express $TS^1 \subset T\mathbb R^2 \cong \mathbb R^4$ as an algebraic variety as follows: $$TS^1 = \{ ((x,y);(u,v)) : x^2+y^2=1 \ \mbox{and} \ xu+yv=0 \}$$ Considering the Jacobian matrix, critical points and critical values shows that $TS^1$ is locally diffeomorphic to $\mathbb R^2$ in a neighbourhood of each of its points.
Taking $xyz$-coordinates on space, and $uvw$-coordinates for the tangent space at the point $(x,y, z) \in S^2 \subset \mathbb R^3_{(x,y,z)}$, I can express $TS^2\subset T\mathbb R^3 \cong \mathbb R^6$ as an algebraic variety as follows: $$TS^2 = \{ ((x,y,z);(u,v,w)) : x^2+y^2+z^2=1 \ \mbox{and} \ xu+yv+zw=0 \}$$ Considering the Jacobian matrix, critical points and critical values shows that $TS^2$ is locally diffeomorphic to $\mathbb R^4$ in a neighbourhood of each of its points.
How do/can I use these deceptively simple equations to prove orientability?
I'm a bit confused. I'm pretty sure $TS^2$ is orientable, both as a vector bundle and as a manifold. As a vector bundle, an orientation is an everywhere nonvanishing section of the determinant bundle ($n$ form), taking the standard area form on the sphere gives an orientation for $TS^2$.
As a manifold, the total space of the tangent bundle of any manifold is orientable. Say you have local trivializations $\phi_i$ of a manifold $M$. Then, the transition functions for $M$ will be $\psi_{ij}(x)=\phi_j\circ\phi_i^{-1}(x)$. These induce trivializations on the total space of the tangent bundle with transition functions $\psi_{ij}(x)$ acting on the "base" coordinates and the linear transformation $d\psi_{ij}(x)$ acting on the "tangent vector" coordinates. The determinant of these change of basis functions is $(det(d\psi_{ij}))^2>0$ regardless of if the base manifold was even orientable. Thus the tangent bundle is always orientable.
Perhaps you mean to say nontrivializable. I don't know any simple proofs with equations, but it boils down to the hairy ball theorem for which you usually use some invariant such as cohomology or the nonvanishing chern class to prove that there's no nonvanishing global sections.