To show that $-1$ is not in $P$, note that if $-1\in P$ then $(-1)(-1)\in P$, which contradicts the fact that if $x \ne 0$ exactly one of $x$ and $-x$ is in $P$.
Next we show that $i\notin P$. Suppose to the contrary that $i\in P$. Then $i^2\in P$, which contradicts the fact that $-1\notin P$.
The same argument shows that $-i\notin P$. This contradicts the fact that if $x\ne 0$, then exactly one of $x$ and $-x$ is in $P$.
Q1. Isn't the inverse of addition destroyed by selecting only the positive members of P?
If you mean that the set $P$ is not closed under taking additive inverses, yes, that's certainly true. For instance, $P$ is not a subgroup of the additive group of the field $F$. Your language ("destroyed") makes this sound like a bad thing, but it isn't. If it helps, the subset $P$ is not the "ordered field" itself, it's the extra structure on the field which allows you to view it as an ordered field, namely $x < y \iff y-x \in P$. Often $P$ is called the positive cone of the ordering, and this language is more suggestive since in both the frozen dairy industry and higher mathematics, cones are not required to be closed under inversion.
Q2. Can we say that this construction is effectively showing that there exists two automorphisms for the field extension $\mathbb{Q}(\sqrt{2})/\mathbb{Q}$?
I would say that the construction is effectively using the nontrivial automorphism $a+ b \sqrt{2} \mapsto a- b\sqrt{2}$ of $\mathbb{Q}(\sqrt{2})$: the logic of it goes in the other direction. Namely, whenever you have an ordering $<$ on a field $K$ and an automorphism $\sigma$ of $K$, you can pull back the ordering by the automorphism to get an ordering $<_{\sigma}$ defined by $a <_{\sigma} b \iff \sigma(a) < \sigma(b)$. In the case when $K = \mathbb{Q}(\sqrt{2})$, this is exactly how we get the second ordering from the first (which itself comes from the ``standard embedding'' into $\mathbb{R}$: in other words, the second ordering $<_\sigma$ on $\mathbb{Q}(\sqrt{2})$
is characterized by $a+b\sqrt{2} >_\sigma 0$ if $a-b\sqrt{2} > 0$.
It is easy to check that in full generality, $<_{\sigma}$ is an ordering on the field $K$. We did not check in general that $<_{\sigma}$ is a different ordering from $<$. In the above example this was clear: e.g. $\sqrt{2}$ is regarded as positive in exactly one of the two orderings. However it is possible for $<_{\sigma}$ and $<$ to coincide: this happens (rather tautologically) exactly when $\sigma$ is not just an automorphism of
the field $K$ but of the ordered field $(K,<)$. Here is what I think is the simplest
example:
Let $K = \mathbb{Q}(t)$ be the rational function field in one variable. Then the automorphism group of $K$ is $\operatorname{PGL}_2(\mathbb{Q}) = \{t \mapsto \frac{at+b}{ct+d} \ | a,b,c,d \in \mathbb{Q}, \ ad \neq bc\}$. Then it makes a nice exercise to show:
(i) There is exactly one ordering $<$ on $\mathbb{Q}(t)$ in which $n < t$ for all $n \in \mathbb{Z}$: one says that the element $t$ is infinitely large.
(ii) The automorphism $t \mapsto t+1$ preserves this ordering.
In general the set $X(K)$ of all orderings on a field $K$ can be naturally endowed with the structure of a topological space: see e.g. $\S 15.8$ of these notes. This is already an interesting structure, and as we saw it carries a natural -- but not necessarily free -- action under the group $\operatorname{Aut}(K)$. (It is easy to see that the action is free when $K$ is a number field. Beyond that I don't know much and would be interested to learn more.)
Q3. What are the practical consequences for having these two distinct ways? If there are many, can you sum up their theme? (Read: why is this a counter example except to the rationals and irrationals, or why is this counter example important?)
In real analysis I don't know of any real consequences of the above considerations, and the above counterexample does seem rather peripheral to me: I was slightly surprised to see it in Gelbaum and Olstead's book.
However, the example is an important one for general culture and perspective. In general, when two objects are conjugate under a group action, it is often awkward to prefer one over the other: that entails a certain symmetry-breaking. An effect of this is to make algebraists view fields more abstractly: it is not fully helpful to view $\mathbb{Q}(\sqrt{2})$ as a subfield of $\mathbb{R}$ because it breaks the symmetry between $\sqrt{2}$ and $-\sqrt{2}$. A more sophisticated (and ultimately more useful) perspective is to start with some more intrinsic definition of
$\mathbb{Q}(\sqrt{2}$) -- e.g. as $\mathbb{Q}[t]/(t^2-2)$ -- and then realize that it has exactly two embeddings into $\mathbb{R}$. Carrying this idea further, a number field $K$ is a finite degree field extension of $\mathbb{Q}$. It is not hard to show that any number field can be embedded into the complex numbers $\mathbb{C}$, so that one could define a number field as a subfield of $\mathbb{C}$ with finite dimension over $\mathbb{Q}$. But this definition is subtly unnatural in number theory: it is much better to keep track of the set of embeddings of $K$ into $\mathbb{C}$...
Best Answer
Take $x <y $, $z>0$. Using (iii), you get $y-x>0$. Now (iv) gives $$(y-x)z>0, $$ which is $$yz-xz>0, $$ and now with (iii) you get $$yz>xz. $$
When $z <0$, use (iii) to get $-z>0$ , and now by the above $$(y-x)(-z)>0, $$ and you unravel this to $$zy <zx. $$
Note that (i) is irrelevant for this. And I would call a field with (i) a "totally ordered" field.