Orientation on Riemann surfaces $\mathcal{X}$ is a Riemann surface and $\mathcal{E}^{(2)}(\mathcal{X})$ is the $\mathbb{C}$-Vector space of all differentiable $2$-forms on $\mathcal{X}$. I want to define the orientation of $\mathcal{X}$:
Why is $\mathcal{X}$ orientable if and only if there exists a  $2$-form $\omega \in \mathcal{E}^{(2)}(\mathcal{X})$ without zeros?
What do we mean with the Standard-Orientation of $\mathcal{X}$?
(Maybe my question is a bit easier if we take  $\mathcal{X}=\mathbb{C}$ as an example.)
 A: It turns out that a complex manifold $X$ (ex. your Riemann surface) has a canonical orientation. Indeed, take any local basis of vector fields $\frac{\partial}{\partial x_i}$, $\frac{\partial}{\partial y_i}$ coming from local holomorphic coordinates $z_i$. Using the CR-equations, you can prove that these bases define a consistent orientation of $X$.
A: This is a general fact about smooth manifolds, you do not need surfaces for this.
Let $M$ be a smooth $n$-dimensional manifold. Suppose there exists a smooth degree $n$ form $\omega$ which does not vanish anywhere. Then, if we write $\omega$ in coordinates $x_i$:
$$
\omega=f(x)dx_1\wedge ... \wedge dx_n.
$$ 
If we change the coordinates to $y_j$:
$$
\omega= g(y)dy_1\wedge ... \wedge dy_n
$$ 
where $g/f=J_h$, Jacobian of the transition map $h$. Now, pick local coordinates such that $f(x)>0$ for every chart. Then the transition map has positive Jacobian, hence, we got an orientation atlas. 
Conversely, suppose that we have an orientation atlas, i.e. one where the transition maps have positive Jacobians. WLOG, we can assume that the atlas is locally finite (since $M$ is finite-dimensional: We can even assume that its multiplicity is $\le n+1$.) Let $(\eta_{\alpha})$ be a partition of unity on $M$ for the corresponding open cover $(U_\alpha)$ of $M$. For each chart $(U_\alpha, \phi_\alpha)$ we let $\omega_\alpha\in \Omega^n(U_\alpha)$ be the pull-back of the form
$$
dx_1\wedge ... \wedge dx_n
$$
under $\phi_\alpha: U_\alpha\to R^n$. Lastly, define
$$
\omega=\sum_{\alpha} \eta_\alpha \omega_\alpha. 
$$
This is the required nowhere vanishing form. To see that $\omega$ has no zeroes, write $\omega$ in local coordinates for some $U_0$. Then, it has the form
$$
\sum_{\alpha} h_\alpha(x) \eta_\alpha(x)dx_1 \wedge ...\wedge dx_n 
$$
and $h_\alpha(x)>0$ for every $x$ (because the transition maps have positive Jacobians, see above). Now, observe that
$$
\sum_{\alpha} h_\alpha(x) \eta_\alpha(x)>
$$
$$
\epsilon \sum_{\alpha} \eta_\alpha(x)=\epsilon>0,
$$
where $h_\alpha(x)\ge \epsilon$ for every $\alpha$ such that $U_\alpha\cap U_0\ne \emptyset$. 
