Why is this inference invalid?

There's no guarantee that the race was run, or that someone won - perhaps everyone wanted to win, but the race was called off due to rain. Perhaps every runner crossed the line at the same time, and the officiator decided that no-one would be declared winner. Perhaps no-one actually signed up for the race - typically if a set $S$ is empty, then a statement like $(\forall x \in S) P(x)$ is vacuously true, but any attempt to say something about a particular element of $S$ will fail to hold any meaning.

Hmm yes, the problem seems to be with this "notation you can't seem to find anywhere" which I've never heard of either. (Though Henning in the comments seems to have pinned it down).

Referring to 'the person who won the race' as an object is odd from the perspective of mathematical logic since it presumes that person exists and is unique. So translating the second sentence into logic is a bit tricky. Since the author has a notation for it, perhaps it's an abbreviation for a first order statement or we're working in a nonstandard system. If it's an abbreviation, it would probably be defined as: $$(|x\,xR)P = ((\exists!x)xR) \wedge ((\forall x) (xR\rightarrow xP)) $$ which is to say that the statement is implicitly asserting that 'the person who won the race' exists.

In the case that there is nobody who won the race, this is false, regardless if $\forall x\, xP$ is true, so that fits with what the author is saying.

The notation is difficult to understand, the way I read it is:

$$\frac{\forall x\;xP}{(|x\;xR)P }$$

is equivalent to:

$$\forall x P(x) \rightarrow \exists x (R(x)\land P(x))$$

Which implies:

$$\forall x P(x) \rightarrow \exists x R(x)$$

And this is indeed not the case as $R(x)$ could never be true, for any x.