By "descending powers of $x$", it means an expansion of the form
$$
\sum_{i=n}^{\infty} a_ix^{-i}
$$
where $n$ can be any positive or negative integer (if the polynomial was of odd degree, then it would require $\sum_{i=n}^{\infty} a_ix^{\frac{1}{2}-i}$). So if we look at the polynomial again, you have
$$
\sqrt{P(x)} = \sqrt{x^6+8x^4−x^3+7x−2} = x^3\sqrt{1+8x^{-2}-x^{-3}+7x^{-5}-2x^{-6}}
$$
We can now perform a taylor expansion on $X=8x^{-2}-x^{-3}+7x^{-5}-2x^{-6}$, so we have $\sqrt{1+X}$.
$$
\sqrt{1+X} \approx 1+\frac{X}{2}+O(X^2)
$$
This gives
$$
\sqrt{P(x)} \approx x^3+4x-\frac{1}{2} + O(x^{-1})
$$
Note that I believe that either you've mis-copied the approximation given in the article, or the article had a typo. It should not be $x^3+7x-\frac{1}{2}$, as the square of that is
$$
(x^3+7x-\frac{1}{2})^2 = x^6 + 14x^4-x^3+49x^2-7x+\frac{1}{4}
$$
which is clearly not a good approximation to $P(x)$. By comparison
$$
(x^3+4x-\frac{1}{2})^2 = x^6 + 8x^4-x^3+16x^2-4x+\frac{1}{4}
$$
which agrees with $P(x)$ in the first three terms, and thus is a good approximation for large $x$.
Seeing as you tagged this sequences and series, I presume you know that you have to be careful when talking about adding up an infinite sequence of terms.
There is an infinite series hidden in your work when you write
$$
\frac{1}{1-x}=1+x + x^2 +x^3 + \cdots,
$$
but this series only converges when $|x|<1$. This sum is the Taylor series for $\frac{1}{1-x}$.
There is an infinite series hidden in your work when you write
$$
\frac{1}{1-x} = -\frac{1}{x}- \frac{1}{x^2}-\frac{1}{x^3}- \cdots,
$$
and this infinite series only converges when $|x|>1$ (or, equivalently, when $\left | \frac{1}{x} \right|<1$). This sum is the Laurent series for $\frac{1}{1-x}$.
So both expressions are correct (for the right values of $x$) but they are not both correct at the same time. One is valid for $|x|<1$ and the other for $|x|>1$.
Edited to add:
Just to clarify the situation a little further:
You are correct when you say that dividing by $-x+1$ should be the same as dividing by $1-x$. There is no problem with this as long as we stop after finitely many steps:
If we stop after three steps in the polynomial long division algorithm, we get
$$
\frac{1}{1-x}=1+x +x^2 +\frac{x^3}{1-x}.
$$
That is, we get an answer of $1+x + x^2$ with a remainder of $x^3$ that has not yet been divided by $1-x$.
Or if we do it the other way, after three steps, we get
$$
\frac{1}{1-x}= - \frac{1}{x}-\frac{1}{x^2}-\frac{1}{x^3} +\frac{1}{x^3}\times \frac{1}{1-x}.
$$
That is, we get an answer of $- \frac{1}{x}-\frac{1}{x^2}-\frac{1}{x^3}$ with a remainder of $\frac{1}{x^3}$ that has not yet been divided by $1-x$.
Let's assume that $x \neq 0$ and $x \neq 1$ so that everything is well-defined. Then
$$
\frac{1}{1-x}=1+x +x^2 +\frac{x^3}{1-x} =- \frac{1}{x}-\frac{1}{x^2}-\frac{1}{x^3} +\frac{1}{x^3}\times \frac{1}{1-x}.
$$
Both ways of doing it do give us the same answer.
So what's the problem? When we try continuing this to infinitely many terms, we need these infinite series to converge to the correct answer. It is not the order $-x+1$ versus $1-x$ that leads to different answers; it is the change from writing down a sum with finitely many terms to writing down a series with infinitely many terms that we have to be careful about.
Best Answer
The reason we usually go with the leading term in polynomial division is because this will make it so that we cancel out the leading term in the numerator. This makes the total degree of what we're working with go down and we stop when we can't go any further.
But, this only works if the leading term of the denominator is smaller than the leading term of the numerator. So we shouldn't be able to apply it to something like $\frac{2x}{x^2+1}$.
But, note that the final term in the denominator has smaller degree than the final term in the numerator. This means that we can always make the final terms equal. Doing it like this will make the total degree of the output larger, hence we get the series expansion for the rational function.
So, in your example, $2x=2x(x^2+1)-2x^3$. Here we're just making the $+1$ part of $x^2+1$ equal to $2x$ and then subtracting out whatever is leftover (some may say the remainder). With this, you can write
$$ \frac{2x}{x^2+1} = 2x - \frac{2x^3}{x^2+1} $$ Note that the power in the numerator increased, so we can do this again, but with $-\frac{2x^3}{x^2+1}$. In this case, $-2x^3=-2x^3(x^2+1) + 2x^5$, and so overall we have $$ \frac{2x}{x^2+1} = 2x - 2x^3 + \frac{2x^5}{x^2+1} $$ And, again, the power in the numerator increased. Since we can increase this without bound, we can continue on to obtain the power series expansion.