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.
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.
Best Answer
You do it when $x$ is small compared to $1$. That is not as silly a remark as it sounds like it is. If you stop the video's approach part way through you get a remainder term just like you might with the top down approach. You could write $$\frac 1{1-x}=1+x+x^2+\frac {x^3}{1-x}$$ This is an algebraic fact, valid for all values of $x$ except $x=1$. When you keep going you get an infinite sum on the right, which is useful as long as it converges. It converges when $|x| \lt 1$. If $|x| \ll 1$ the sum converges quickly and you can decide when the error committed by truncating it is acceptable.
In some problems you know that $x$ is small, in which case this is quite useful.
If the division is going to come out even, you can do it either way around. Taking your example $$\require{enclose} \begin{array}{r} -7 + 5x \hspace{4pt}+x^2\hspace{33pt} \\ -1+2x \enclose{longdiv}{\hspace{10pt}7-19x+9x^2+2x^3\hspace{4pt}} \\ \underline{7-14x}\hspace{30.5pt} \\ \hspace {30 pt}- 5x+9x^2\hspace{4pt} \\ \underline{- 5x+10x^2}\hspace{21.5pt} \\ -x^2 + 2x^3\hspace{4pt} \\ \underline{-x^2+2x^3}\hspace{0pt} \\ 0\hspace{4pt} \\ \end{array}$$ where I can't get the spacing as nice as you did, but it gives the same result.