When $f$ is a smooth function, then
$$\partial f = \sum \frac{\partial f}{\partial z^i} dz^i, \ \ \overline\partial f = \sum \frac{\partial f}{\partial \bar z^i}d\bar z^i. $$
In general if $\omega $ is a $(p, q)$ form, locally written as
$$\omega = \sum_{|I|=p,|J| = q} f_{IJ} \ dz^I\wedge d\bar z^J,$$
then
$$\partial \omega = \sum_{|I|=p,|J| = q} \partial f_{IJ} \wedge dz^I\wedge d\bar z^J$$
and similar for $\bar \partial \omega$. Note that $d = \partial +\bar\partial$ (direct checking), so
$$0 = d^2 = (\partial + \bar\partial)(\partial +\bar \partial) = \partial ^2 + (\partial \bar\partial + \bar\partial \partial) + \bar\partial^2,$$
this implies that
$$\partial^2 = \partial \bar\partial + \bar\partial \partial = \bar\partial^2 = 0.$$
It is not true that $\partial \bar\partial =0$ though, as shown in the other answers.
This is the definition of $\partial, \bar\partial$ when acting on forms. But if you want the definition on current, it's just the dual: Given a current $T$, then it is just a linear functional on the space of $k$-forms. Define
$$\partial T(\omega) := -T(\partial \omega).$$
You might find more information in the book "Principal of Algebraic Geometry" by Griffiths and Harris, Chapter 3.