I think something more needs to be said. The problem is that an orientation preserving diffeomorphism may be orientation reversing on a subspace. For example, Consider the map $h:\mathbb{R}^3\rightarrow\mathbb{R}^3$ given by $h(x,y,z) = (-x,-y,z)$. This is orientation preserving since the determinant of the Jacobian is 1.
However, the yz plane has its orientation reversed. This is because if you restrict to these points, in local coordinates, you get that $(y,z)$ is mapped to $(-y,z)$, which is clearly orientation reversing. If you prefer a compact example, use the unit circle in the yz plane.
I'm not sure if the fact that the sphere is codimension 1 in $\mathbb{R}^{n+1}$ rules this kind of behavior out or not.
Finally, here is an argument that shows the antipodal map is orientation preserving when $n$ is odd.
Lemma 1: Whether or not a diffeomorphism is orientation preserving or reversing can be checked at a single point (at least if the oriented manifold is connected).
Proof: Given a diffeomorphism $f:M\rightarrow M$, the map from $M$ to $\{-1,1\}$ given by taking the sign of $\det(df(p))$ can be shown to be continuous using local coordinates. (The sign is never $0$ since $f$ is a diffeomorphism). If $M$ is connected, this implies the map is constant.
Lemma 2: If $F:M\times[0,1]\rightarrow M$ is smooth and $f_t(p) = F(p,t)$ is a diffeomorphism for each fixed $t$, then all $f_t$ are orientation preserving or they are all orientation reversing.
Proof: By Lemma 1, we can focus on a single point $p$. Then the function taking $[0,1]$ to the sign of $df_t(p)\in\{-1,1\}$ can be shown to be continuous using local coordinates, hence it's constant.
Lemma 3: When $n$ is odd, there is a smooth map $F:S^{n}\times[0,1]\rightarrow S^n$ for which $f_0$ is the identity map and $f_1$ is the antipodal map.
Proof: For $p = (p_1,...,p_{2n})$, Let $F(p,t) = (\cos(\pi t) p_1 + \sin(\pi t)p_2, -\sin(\pi t) p_1 + \cos(\pi t) p_2,... )$, so essentially do a rotation on each pair of coordinates. This is a diffeomorphism for each $t$ because the inverse map is given by rotation each pair of coordinates the opposite direction for the same time.
Finally, $f_0(p) = F(p, 0) = (p_1,...,p_{2n})$ so $f_0 = Id$ and $f_1(p) = F(p,1) = (-p_1,...,-p_{2n}).$ so $f_1$ is the antipodal map.