The general procedure is correct. And the details look pretty good too. I could now check the details, it would not be difficult.
Instead, I will do the problem in a somewhat different way, which in general is more efficient. It will turn out that a sign error crept into your calculation.
We have $f(x)=-\frac{1}{x}-\frac{1}{x^2} + \frac{1}{x-1}$ and want to express $f(x)$ as a sum of powers of $x-2$. It is useful, though not necessary, to let $y=x-2$. Then $x=y+2$. Substituting for $x$, we obtain $f(y+2)=g(y)= -\frac{1}{y+2}-\frac{1}{(y+2)^2} + \frac{1}{y+1}.$
We want to express $g(y)$ as a sum of powers of $y$. Let's start with the easiest part, $\frac{1}{1+y}$.
It would be a shame to do a whole lot of differentiating when we already know the power series expansion of $1/(1+y)$. Or at least we certainly know the power series expansion of $1/(1-z)$, and then we can put $y=-z$. Thus $\frac{1}{1+y}=1-y+y^2-y^3+\cdots=\sum_0^\infty (-1)^ny^n\qquad\qquad\text{(Term $1$)}$
That was easy. Let's go on to the next easiest term, $1/(y+2)$ (I know there should be a minus sign in front, will take care of it later). We have $\frac{1}{2+y}=\frac{1/2}{1+y/2}.$ It would be a shame not to use the fact that we know the power series expansion of $1/(1-z)$. We get $\frac{1}{2+y}=\frac{1}{2}\sum_0^\infty (-1)^n\frac{1}{2^n}y^n=\sum_0^\infty \frac{(-1)^n}{2^{n+1}}y^n,$ and therefore $-\frac{1}{2+y}=\sum_0^\infty \frac{(-1)^{n+1}}{2^{n+1}}y^n.\qquad\qquad\text{(Term $2$)}$
Finally, we want the expansion of $1/(2+y)^2$ in powers of $y$. We have just obtained the expansion of $1/(2+y)$. Note that $1/(2+y)^2$ is (almost) the derivative of $1/(2+y)$. To be precise, it is the negative of the derivative of $1/(2+y)$. So let us differentiate the series we obtained for $1/(2+y)$ term by term. We find $-\frac{1}{(y+2)^2}=\sum_0^\infty \frac{(-1)^{n} n}{2^{n+1}}y^{n-1}=\sum_0^\infty \frac{(-1)^{n+1} (n+1)}{2^{n+2}}y^{n} \qquad\qquad\text{(Term $3$)}$
Now it is just a matter of adding Terms $1$, $2$, and $3$ together. I get, replacing $y$ by $x-2$, the following: $\sum_0^\infty (-1)^n\left(1-\frac{n+3}{2^{n+2}}\right)(x-2)^n.$
Comment: When we are calculating power series expansions, it is nice to avoid all those differentiations, by recycling standard expansions.