Intuition for why the difference between $\frac{2x^2-x}{x^2-x+1}$ and $\frac{x-2}{x^2-x+1}$ is a constant?

Would you be surprised that the difference of $\dfrac{2x^2+x+1}{x^2}$ and $\dfrac{x+1}{x^2}$ is $2$?

It is just a bit of clever disguise. Take any polynomial $p(x)$ with leading term $a_n x^n$.

Now consider $$\frac{p(x)}{p(x)}$$ This is clearly the constant $1$ (except at zeroes of $p(x)$).

Now separate the leading term: $$\frac{a_n x^n}{p(x)} + \frac{p(x) - a_n x^n}{p(x)}$$

and re-write to create the difference:

$$\frac{a_n x^n}{p(x)} - \frac{a_n x^n - p(x)}{p(x)}$$

Obviously the same thing and hence obviously still $1$ but the first has a degree $n$ polynomial as its numerator and the second a degree $n - 1$ or less polynomial.

Similarly, you could split $p(x)$ in many other ways.

I'm not sure anyone is speaking to your observation that the two numerators have different degrees.

Let's flip this around the other way:

\begin{align*} \frac{x-2}{x^2-x+1} + 2 &= \frac{x-2}{x^2-x+1} + 2\frac{x^2-x+1}{x^2-x+1} \\ &= \frac{2x^2 -x}{x^2-x+1} \text{.} \end{align*} That is, we started with a thing having a linear numerator and added a constant to it. But when we brought the constant to have a common denominator, it picked up a degree two factor. Then the addition was forced to produce a degree two sum.

To sum up, in the context of rational functions, when you add constants, you are adding polynomials having the degree of the denominator to the polynomials in the numerators. So constants effectively have "degree two in the numerator" in your example.