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.
The resulting object will be (assuming everything's compact, say...) an orientable manifold, but not an oriented one. If you want it to be oriented, with the orientation matching that of $M_1$, say, then the map $f$ should be orientation reversing (which may not be possible without reversing the orientation on $M_2$, and hence the induced orientation on $\partial M_2$).
If you think hard about two unit disks in the plane, joined to make a sphere, you'll see what I mean.
post-comment edits
A manifold $M$ is a topological space $X$ together with a (maximal) atlas $A$ that satisfies certain rules that are well-known to the questioner, so I won't write them out. An oriented manifold is the same thing, together with a subset $S$ of the atlas that consists of charts that are declared to be "orientation preserving" (and which, when they have overlapping domains, satisfy something that's the equivalent (in the smooth case) of the transition function's derivative having positive determinant on the overlap).
To any oriented manifold, $(X, A, S)$ there's an associated manifold $(X, A)$ gotten by ignoring the orientation.
If we glue oriented manifolds $M_1$ and $M_2$ together as in the question and if their boundaries are each connected and $f$ is an orientation-reversing homeomorphism between them, then $M = M_1 \cup_f M_2$ is orientable, and indeed, the orientation can be made consistent with that of $M_1$ and $M_2$, in the sense that both $M_1$ and $M_2$ are embedded in $M$, and the embeddings are orientation-preserving. The glued up manifold is orientable, but until we pick an orientation, it's just a manifold! And the homeomorphism class of that manifold is independent of the orientations of $M_1$ and $M_2$. That's the claim that Michael Albanese, in the comments, seems willing to believe. Let's look at the orientation preserving case now.
If $f$ happens to be orientation-preserving, then we can reverse the orientation on $M_2$ to get a different oriented manifold, $M_2'$, and a map $f' : \partial M_1 \to \partial M_2': x \mapsto f(x)$ which is exactly the same as $f$, except that the target oriented-manifold is now $M_2$ with the other orientation. Note, too, that $M_2'$ and $M_2$ are homeomorphic as manifolds, by the identity map, but are not necessarily homeomorphic by an orientation-preserving homeomorphism.
If we now build $M' = M_1 \cup_{f'} M_2'$, it's homeomorphic to $M$, but not necessarily oriented-homeomorphic. It's orientable, and with the right orientation, we find that $M_1$ and $M_2'$ are both embedded in it by orientation-preserving embeddings.
On the other hand, it's not true that $M_2$ is embedded in it by an orientation-preserving embedding.
What IS true is that $M_2$ is embedded in it as a topological submanifold-with-boundary.
In the case Michael asked about, where $\partial M_1 = \partial M_2 = S^3$, and $M_1$ and $M_2$ are both $\Bbb CP^2$, we get that
$$
M' = \Bbb CP^2 \# \overline{\Bbb CP^2}
$$
where the "equality" here is an orientation-preserving homeomorphism (basically just "inclusion"). But it's also true that $M'$ is homeomorphic (again by inclusion) to
$\Bbb CP^2 \# \Bbb CP^2$, by a non-orientation-preserving homeomorphism.
Best Answer
Actually, let's even forget about volume forms.
Start with the manifold ${\bf R}^{N}$, regard it as a vector space (in the future, as a tangent space at a point of a manifold), and choose a basis: $$ {\bf E} = \{{\bf e}^{k} \; : \; k = 1, 2, \ldots, N\} $$ (in the future, a frame on the manifold), with the qualification that the basis is ordered (say, by the indices $k$). (For $N$, the order ${\bf e}^{1} = (1, 0), {\bf e}^{2} = (0, 1)$ corresponds to what we know colloquially as the counterclockwise orientation; to help the visual, traverse the points points: ${\bf 0} = (0,0), {\bf e}^{1} = (1, 0), {\bf e}^{2} = (0, 1)$, in that order).
A linear transformation $A$ which transforms one basis (frame) to another will, therefore, be invertible, hence has a nonzero determinant. The transformation with a positive determinant is said to $\textit{preserve orientation}$ with a negative determinant, to $\textit{change orientation}$. (Test this on the 2-D example I gave above.)
Two frames have the same orientation if one is transformed into the other by a transformation with a positive determinant. This establishes an equivalence relation on the set of frames, with two equivalence classes. These classes are called orientations.
Hope this sheds light on it.
A source: Mathematical Analysis II, V.A. Zorich