In the book of D. Chillingworth titled Differential Topology with a view to applications, at page 141, the author argues that
[...] $\mathbb{P}^2$ can be described equivalently as the set of all paris of antipodal points in $S^2$, [...] $\mathbb{P}^2$ is the quotient space $S^2 / R$ where $R$ is the equivalence relation "lying on the same line through the origin".
[...] The easiest way to see $\mathbb{P}^2$ is non-orientable is to think a coin heads uppermost near the north pole on $S^2$ being slid around the near the south pole and then being projected back to near the north pole directly by the antipodal map: it ends up tails uppermost.
However, I cannot understand the given intuitive argument about "projecting the coin by the antipodal map"; while we are sliding the coin towards the south pole, it is still head uppermost, and since the north and the south poles are the same points, being in the latter is the same as being in the former, so I don't see why the antipodal map needs to "flips" the coin.
Question:
What is the intuitive argument that the author is trying to convey ?