I was reading through the proof that the tangent bundle is orientable here and was stuck on one part: why (at the bottom of the 7th page) does taking $d(\overline{\phi} \circ \overline{\psi}^{-1})$ not change the second half of the coordinates? I would expect something like $d(d(\phi \circ \psi^{-1}))$ but instead nothing changes? Everything else makes sense but perhaps I'm missing something fundamental about tangent bundles. I've tried to write things out in coordinates, but I get second partial derivatives in the second half when I do that, which doesn't match what they're saying...
For example, in the 1-dimensional case, we have coordinates $(x_1, u_1 \frac{\partial}{\partial x_1})$ on $TM$. So applying $d$ (writing $f = \phi \circ \psi$) gives $(\frac{\partial f}{\partial x_1} + \frac{\partial f}{\partial u_1}, \frac{\partial^2f}{\partial x_1^2} \frac{\partial}{\partial x_1} + \frac{\partial^2 f}{\partial x_1 u_1} \frac{\partial}{\partial u_1})$ and then I don't see how this is supposed to be the same as their expression.
You are correct, the note is missing one part of the differential. To simplify notations, let $x = (x^1, \cdots, x^n)$, $y = (x^{n+1}, \cdots, x^{2n})$ and $f = \phi \circ \psi^{-1}$. The mapping concerned is
$$ F(x, y) = (f(x), d_xf(y)).$$
It's differential is given by
$$d_{(x,y)} F (u,v) = (d_xf (u), \langle d_x(d_xf(y)), u\rangle + d_xf(v))$$
instead of $(d_xf(u), d_xf(v))$ as in the note (The main point is that when you differentiate $d_xf (y)$ with respect to $y$, you got $d_xf$ since $d_xf(y)$ is linear in $y$). As a matrix, $dF$ is given by
$$d_{(x,y)} F(u,v) = \begin{bmatrix} u & v \end{bmatrix}\begin{bmatrix} d_x f & d_x(d_xf(y)) \\ 0 & d_xf \end{bmatrix} $$
and so $\det d_{(x,y) }F) = (\det (d_xf))^2$. Hence the argument is still the same.