This is the explanation from Jänich on why the antipodal map $\tau:S^n \rightarrow S^n$ reverses orientation iff $n$ is even:
For every $x\in S^n$, the differential of the diffeomorphism $-\text{Id}:D^{n+1}\rightarrow D^{n+1}$ takes the outward normal $\vec{N}(x)$ at $x$ to $\vec{N}(-x)$ at $-x$, so the diffeomorphism reverses the orientation of the boundary exactly when it reverses the overall orientation of $D^{n+1}$, and it obviously does the latter iff $n$ is even.
All I understand is this:
But why does that mean that the diffeomorphism reverses the orientation of the boundary exactly when it reverses the overall orientation of $D^{n+1}$? Is this because if I reverse the boundary I need to reverse the overall orientation to keep the orientation convention? Something like two wrongs make a right? (and by a wrong I mean reverse the orientation of a single vector, or do I need to reverse all the vectors that are not the normal?) Why is it obious that $n$ has to be even?
This is very hard for me to picture, is there any other way to see it?
The first problem is that your picture is incorrect. The correct picture is:
However, that picture alone doesn't explain what is going on, as picture looks the same in every dimension. In order to see what is going on, you need to look at all axes, not just one. So I added the second, tangent, axis in red. Note that $-Id$ also reverses its direction. But it still lies in the same relation to the normal axis both before and after the operation. The tangent axis is clockwise from the normal axis. If I rotate the circle by $\pi$, we get right back to the original picture.
Contrast this to $S^2$ in $\Bbb R^3$. Then all three axes reverse. This swaps the traditional right-handed coordinate system to a left-handed one. Simply rotating the sphere to bring the antipodal point back to the original position does not align to reversed axes back to their original positions. The normal rotates around to the same position again, but while by choosing the appropriate rotation, you can bring one of the two tangent directions around to match its previous alignment, the other tangent direction will always be reversed to its original alignment.
Stick up your right and left hands with thumb, index and ring fingers at right angles. Notice how you can align the thumbs, index and ring fingers on each hand to all three point in opposite directions at the same time? They correspond to the three directions before and after applying the negation map. See that no matter how you rotate your hands, you can never get all three to point in the same directions on each hand. (If you could, there would be no distinction between "right" and "left".)
This reversal of orientation of axes by negation only occurs in an odd number of dimensions. In an even number, you can always pair them up and rotate them as in the picture above back into their original orientation. In an odd number of dimensions, you always have one dimension left over who has no "buddy" that it can pair up with to rotate back to its original orientation. So while everyone else gets to go home, it is left facing in the opposite direction.
To apply this to $S^n$, he embeds $S^n$ into $\Bbb R^{n+1}$. There, the antipodal map on $S^n$ is the restriction of negation in $\Bbb R^{n+1}$. When that negation is applied, all directions reversed. If $n$ is odd ($n + 1$ is even), they can be rotated back to match the original directions. If $n$ is even ($n+1$ is odd), one direction cannot be aligned back up by rotations. If all the tangent vectors are aligned, then it will be the normal direction that has to switch. Thus the orientation reversed.