This is an elaboration on my comments on the question, and on Jason DeVito's answer.
Firstly, what does it mean to choose an orientation on a manifold $M$?
One way to think of it is that an orientation is a collection of charts covering $M$ such that the Jacobians of the change-of-coordinate map on all overlaps have positive determinants.
Another way to think of an orientation is that we have to choose an orientation
for $TM_p$ for each $p \in M$, with the property that given any $p \in M$,
there is some chart $U$ containing $p$, with local coordinates $x_1,\ldots,x_n$,
such that the orientation on $TM_q$ is the one that contains
the basis $\partial_{x_1}, \ldots, \partial_{x_n}$,
for each $ q \in U$. (Recall that an orientation on a vector space over
$\mathbb R$ is a collection of bases such that all change of basis matrices
have positive determinant.)
It's not hard to check that these two notions coincide. Indeed, given
a colllection of charts as in the first definition,
we can define an orientation on the tangent space $TM_p$ for each $p\in M$ as follows: if $p \in U$ for one chart $U$ in our given collection, and if $x_1,\ldots,x_n$ are the local coordinates on $U$, then we define the orientation on $TM_p$ to be the one containing the basis
$\partial_{x_1},\ldots,\partial_{x_n}$. Note that our assumption on the transition maps on the overlaps of the charts means that this really does give a well-defined orientation on the vector space $TM_p$ for each $p$. By construction
the resulting set of orientations on the tangents spaces $TM_p$ satisfies the conditions of 2.
Conversely, given a set of orientations on the $TM_p$ as in definition 2,
consider the set of charts $U$ whose existence is guaranteed by 2; this collection of charts evidently satisfies the conditions of definition 1.
For the sphere, there is a standard way to choose an orientation: fix a unit normal
vector field $\mathbb n$ on $\mathbb S^n$, either the inward pointing normal or the outward pointing normal. Also fix an orientation on $\mathbb R^{n+1}$ as a vector space.
If $p \in \mathbb R^{n+1}$, then $T\mathbb R^{n+1}_p \cong \mathbb R^{n+1}$ canonically, and so we get an orientation on $T\mathbb R^{n+1}_p$ for each $p$.
(A slightly more long-winded way to describe what I just did, which might nevertheless be helpful, is: I am using $\mathbb R^{n+1}$ as a global chart on itself, and hence defining an orientation on $\mathbb R^{n+1}$ as a manifold as in definition 1. I am then using the procedure described above, of going from 1 to 2, to get an orientation on each $T\mathbb R^{n+1}_p$.)
Now for each $p \in \mathbb S^n$, define an orientation on $T\mathbb S^n_p$
such that the induced orientation on $T\mathbb S^n_p \oplus \mathbb R\mathbb n = T\mathbb R^{n+1}_p$ (induced orientation meaning that we add $\mathbb n$ to any positively oriented basis of $T\mathbb S^n_p$ so as to get a basis for
$T\mathbb R^{n+1}_p$) coincides with the given orientation on $T\mathbb R^{n+1}_p$.
Now that we have fixed on orientation on $\mathbb S^n$, we are finally in a position to make a Jacobian computation to compute whether $f$ preserves or reverses orientation.
As noted by the OP, $Dh(p)$ has determinant $(-1)^{n+1}$ for any point $p$.
On the other hand, $Dh$ takes the unit normal $\mathbb n(p)$ to the unit normal
$\mathbb n(f(p))$. (Draw the picture!)
In other words, when we consider $Dh(p): T\mathbb R^{n+1}_p \to
T\mathbb R^{n+1}_{f(p)}$, and we decompose this into the direct sum of $Df(p): T\mathbb S^n_p
\to T\mathbb S^n_{f(p)}$ and the map $\mathbb R \mathbb n(p) \to \mathbb R \mathbb n(f(p))$ induced by $Dh(p)$, the latter map has the matrix $1$ with respect to the bases $\mathbb n(p)$ in the source and $\mathbb n(f(p))$ in the target. Thus $Df(p)$ also has determinant $(-1)^{n+1}$ with respect to the
positively oriented bases on its source and target.
Thus $f$ is orientation reversing/preserving according to whether $n$ is even/odd.
First you should look for the Euclidian case, supose that $U$ and $V$ are open connected sets on $\mathbb{R}^n$ and $f:U \to V$ is diffeo. So by definition $f$ preserve orientation in $x$, if the linear transformation $df_x: \mathbb{R}^n \to \mathbb{R}^n$ preserve orientation, or in a equialent way $\mathrm{det}(df_x)>0$. Now lets proof that if exists $x_0 \in U$ s.t. $f$ preserve orientation on $x_0$ then $f$ preserve orientation in every point in $U$, but as you should know the map $d: U \ni x \mapsto \mathrm{det}(df_x) \in \mathbb{R}$ is smooth, now note that because $f$ is a diffeo. the map $d$ is never zero, so if exists $y \in U$ s.t $d(y)<0$ the facts that $d(x_0)>0$ and $U$ is connected implies that should exists $y' \in U$ s.t. $d(y')=0$, absurd so $f$ preserve orientaion on every point.
Now for the general case $f:M \to N$. Define the set:
$$A=\{x \in M : \mbox{$f$ preserve orientation on x}\},$$
we will prove that $A$ is closed, open and non-empty, so $A=M$. First $A$ is non-empty by hypothesis. Now lets proof that is open, let $x_0 \in A$, so by definition exists charts $\phi:U\subset M \to U' \subset \mathbb{R^n}$ and $\psi:V\subset N \to V' \subset \mathbb{R}^n$ s.t the map $\bar{f}: U' \to V'$, given by $ \bar{f}=\psi^{-1} \circ f \circ \phi$, satisfies $\mathrm{det}(d\bar{f}_{\phi(x_0)})>0$. So by the Euclidian case exists a connected nbhd $W$ of $\phi(x_0)$ s.t $\bar{f}$ preserve orientation on $W$, so by definition $f$ preserve orientation for every point in $\phi^{-1}(W)$. By definition of $A$ we have that $\phi^{-1}(W) \subset A$, that is, $A$ is open as claimed. The proof that $A$ is closed is pretty similar.
Best Answer
Note that $a_*:T_pM\to T_{a(p)}M$ is a map between different tangent spaces. Without specifying charts, the statement that the local representative of $a_*$ at some point $p$ is $-I_{n}$ is vacuous: with suitable choices of local coordinates around $p$ and $a(p)$, you could make the local representative of $a_*$ whatever you wanted.
To make the orientation of local representatives meaningful, you would need to choose consistently oriented charts around $p$ and $a(p)$. If you do, you will find that the local representatives have appropriately signed determinant.
Alternately, a common approach for this problem is to work with the standard embedding $S^n\subset\mathbb{R}^{n+1}$ since both the antipodal map $a$ and the standard orientation form on $S^n$ are easy to work with in this setting. Since $\mathbb{R}^{n+1}$ is covered by a single chart (which, of course, is consistently oriented with itself), we can work in coordinates and the determinant of local representatives of $a$ are meaningful. It will take a bit more work to restrict the results to $S^n$, of course.