For Lebesgue integrals, I know how to approximate $\int_{A} f(x)dx$ for some measurable set $A\subset \mathbb{R}^{n}$, I just decompose $A$ into measurable subsets $A_{i}$ and then make some finite "Riemann sum" $\sum f(x_{i})\mu (A_{i})$ where $x_{i}$ is a well-chosen point in $A_i$.
For integration over forms, I guess that I cannot anymore do this, since now I need to deal with orientation and stuff. Basically, I cannot just treat $dx\wedge dy \wedge dz$ as $dxdydz$. How do I proceed then to obtain a numerical approximation of $\int_{A} f(x,y,z) dx\wedge dy \wedge dz$ for some $A\subset \mathbb{R}^{3}$ ?
Underlying an integral of a differential form is the Lebesgue integral. In fact, using the standard orientation of $\mathbb{R}^3$, $$ \int_A f(x,y,z)\,dx\wedge dy\wedge dz = \int_A f(x,y,z)\,dV, $$ where $dV$ is the standard Lebesgue measure on $\mathbb{R}^3$.
Differential forms are needed for defining integrals whose value does not depend on the coordinates used. They were developed in order to define an integral over a manifold (on which there is no coordinate-independent way to integrate a function).
But to calculate an integral of a differential form, you have to write the differential form with respect to coordinates. For example, if $\omega$ is a $3$-form on $A$, then with respect to standard coordinates, it can always be written as $$ \omega = f(x,y,z)\,dx\wedge dy\wedge dz.$$ Therefore, $$ \int_A \omega = \int_A f(x,y,z)\,dx\wedge dy\wedge dz = \int_A f(x,y,z)\,dV. $$ Note that the orientation plays a role when writing $\omega$ in this form. For example, if $$ \omega = dy\wedge dx \wedge dz, $$ then $$ \int_A \omega = \int_A dy\wedge dx \wedge dz = -\int_A dx\wedge dy\wedge dz = -\int_A dV. $$