Cover of a sheaf We define $U_1:=\mathbb{P}^1\setminus\left\lbrace \infty\right\rbrace$ and $U_2:=\mathbb{P}^1\setminus\left\lbrace 0\right\rbrace$. 
$\Omega$ is the sheaf of holomorphic $1$-forms.
How can I understand that $U=(U_1,U_2)$ is a Leray cover of $\mathbb{P}^1$ relating to the sheaf $\Omega$?
I think therefor I have to show that $H^1(U_1,\Omega|_{U_1})=H^1(U_2,\Omega|_{U_2})=0.$
 A: The $U_i$ are both isomorphic to $\mathbb{C}$, and their intersection is $\mathbb{C}^*$. If you know these two spaces have no higher cohomology in $\Omega,$ then you are done. An analytic proof of this follows, for instance, from the $\bar\partial$-Poincare lemma, one version of which you cite in your second comment. Your idea won't quite work, because the $H^1$ you want to compute actually isn't described by $(1,0)$-forms, though, but rather $(1,1)$ forms. This is Dolbeault's theorem relating Dolbeault cohomology to sheaf cohomology.
If you don't know the theorem, you can do it directly: there's a short exact sequence of sheaves $0\to\Omega\to \mathcal A^{1,0}\stackrel{\bar\partial}{\to} \Omega^{1,1}\to 0$, where $\Omega^{1,1}\subset \mathcal A^{1,1}$ is the $\bar\partial$-closed forms, which coincide with the $\bar\partial$ exact ones by the $\bar\partial$ lemma. So writing the long exact sequence and using fineness of $\mathcal A^{p,q}$, $H^1(U_1,\Omega)$ coincides with $\Omega^{1,1}/\bar\partial \mathcal A^{1,0}$, which vanishes, again by the $\bar\partial$ lemma. The same goes for $U_2,U_1\cap U_2$.
EDIT: So now we must compute the cohomology of the cover $U$. $C^0(U)=\{(\xi_1,\xi_2):f\in \Omega(U_1),g\in\Omega(U_2)\}$. But as you've said this means $\xi_1=f(z)dz$ and $\xi_2=g(w)dw$ where $f(z)=\sum_{k=0}^\infty f_k z^k$ is holomorphic in $z$, and similarly for $g$. The coboundary map is $\delta(fdz,gdw)=(f-g)dz$, since $C^1$ just consists of holomorphic 1-forms on $U_1\cap U_2$.. To compute this we must write $g$ in the same coordinate as $f$, so $w=1/z$ and $dw=-(z)^{-2}dz$. That is, on the intersection we write $gdw=g(1/z)d(1/z)=-z^{-2}\sum_{k=-\infty}^0 g_k z^k$dz=$\sum_{k=-\infty}^0-g_kz^{k-2}dz$. So $(f-g)dz=\sum_{k=0}^\infty f_k z^k-\sum_{k=-\infty}^0g_kz^{k-2} dz$. This is zero if and only if both $f$ and $g$ are zero, since none of the exponents agree in the two terms, which shows $H^0(U,\Omega)=0$. 
For $H^1$ we only need the image of $\delta$; since there are no 3-way intersections, the kernel is all of $C^1$. As shown above $(f-g)dz$ has no $z^{-1}$ term. Now $C^1(U,\Omega)=\{hdz\}$ where $h$ is holomorphic on $\mathbb{C}^*,$ that is, of the form $\sum_{k=-\infty}^\infty h_k z^k$, and we can write such an $h$ as $(f-g)dz$ if and only if $h_{-1}=0$. Thus the quotient $C^1/\delta C^0$ has representatives $h_{-1}z^{-1}dz$ where $h_{-1}$ ranges over $\mathbb{C}$ and $H^1(U,\Omega)=\mathbb{C}$.
