Let first state the version of Stokes theorem I am looking at
For any smooth $(n-1)$-form $\omega$ with compact support on the oriented $n$-dimensional manifold $M,$ $\int_M d\omega = \int_{\partial M} \omega$.
I never learn to know how to use this properly. One specific problem I have at hand is the following:
$f$ is a smooth function defined on the unit disk $\Delta,$ $$ \eta = \frac{1}{2 \pi i} \frac{f(w)\,dw}{w-z}, $$ $$ d\eta = - \frac{1}{2\pi i} \frac{\partial f(w)}{\partial\bar w} \frac{d \omega \wedge d \bar \omega}{w-z}, $$ let $\Delta_{\varepsilon}$ be the disc of radius $\varepsilon$ around $z$. Then by stokes theorem:
$$ \frac{1}{2\pi i}\int_{\partial \Delta_{\varepsilon}} \frac{f(w) \, dw}{w-z} = \frac{1}{2\pi i} \int_{\partial \Delta}\frac{f(w) \, dw}{w-z} + \frac{1}{2\pi i} \int_{ \Delta - \Delta_{\varepsilon}} \frac{\partial f(w)}{\partial\bar w} \frac{d \omega \wedge d \bar \omega}{w-z}.$$
Can someone explain to me in detail how does the Stokes theorem apply here?
The orientation is important in the proof of the stokes theorem since we want to choose a set of oriented atlas. But I do not see how do we see the orientation in real applications.
Consider applying Stokes theorem on the annulus $\Delta-\Delta_{\epsilon}$, using your differential form $\eta$:
$\int_{\Delta-\Delta_{\epsilon}} d\eta = \int_{\partial(\Delta-\Delta_{\epsilon})} \eta$ .
The boundary $\partial(\Delta-\Delta_{\epsilon})$ consists of 2 circles, the outer one being the circumference of $\Delta$ and the inner one being the circumference of $\Delta_{\epsilon}$. Like Steve have commented, these 2 circles are oppositely oriented and so the above equation becomes
$\int_{\Delta-\Delta_{\epsilon}} d\eta = \int_{\partial\Delta} \eta -\int_{\partial\Delta_{\epsilon}} \eta$ .
Then plug in your $\eta$ and $d\eta$ to get what you want to show.