Let $S=\{0, 1, \infty\} \subset \mathbb{P}^1$ and let $\Omega^1_{\mathbb{P}^1}(\log S)$ be the line bundle of logarithmic differentials along $S$. Consider the form $$ \omega=\frac{dx}{x}+\frac{dx}{x-1} $$ which is an element in $H^0(\mathbb{P}^1, \Omega^1_{\mathbb{P}^1}(\log S))$ with residues $1$, $1$ and $-2$ at $0$, $1$ and $\infty$.
What is the divisor of $\omega$ regarded as (rational) section of $\Omega^1_{\mathbb{P}^1}(\log S)$?
Writing $$ \omega=\frac{2x-1}{x(x-1)}dx, $$ I would simply say that $$ \mathrm{div}(\omega)=[1/2], $$ since $x=1/2$ is the point where the numerator has a simple zero. But I am a bit confused about the difference between considering $\omega$ as a rational section of $\Omega^1_{\mathbb{P}^1}$ or $\Omega^1_{\mathbb{P}^1}(\log S)$.
Is it OK that $\mathrm{div}(\omega)$ has degree 1? Am I just computing the Chern class of $\Omega^1_{\mathbb{P}^1}(\log S)$?
Can someone help?
The degree of $\Omega^1(\log S)$ is equal to $\deg \Omega^1 + \deg S = - 2 + 3 = 1,$ and so it is natural that $\div(\omega)$ has degree $1$ when thought of as a section of $\Omega^1(\log S)$.
In particular, your computation looks fine to me.