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.
Best Answer
It is not true that if a linear map $f:V\to W$ has a one sided inverse, then $\dim V=\dim W$ nor that $f$ is an isomorphism.