You can interpret the differential 2-form $dz_1 \wedge dz_2$ on $\mathbb{C}^2$ also as a 1-form on $\mathbb{C}^2$ with value in $T^*\mathbb{C}^2$. As such, you would be able to integrate such a form on 1-submanifolds in order to get a 0-form on $\mathbb{C}^2$ with value in $T^*\mathbb{C}^2$, i.e. a usual 1-form.
(To get a sense of what this means, recall from calculus classes the integration of the velocity vector in order to get the displacement of a point particle. How did you defined such an integration, if not as the 'usual' Riemann integration of a vector-valued function instead of a scalar one? Here, over the point $q \in \mathbb{C}^2$, the 1-form takes value in the vector space $T^*_q \mathbb{C}^2 \simeq \mathbb{C}^2 \simeq \mathbb{R}^4$, so the integration of such a 1-form is not so very different from integration in calculus.)
Somewhat put differently, imagine you were to integrate $dz_1 \wedge dz_2$ over the 2-manifold $\gamma_1 \times \gamma_2$ where $\gamma_j \subset [z_j=0]$. In practice, you would compute this integral through two iterated integrals, first performing the integration over each slice $\gamma_1 \times \{p\}$ with $p \in \gamma_2$, then the integral over $\gamma_2$. In some respect, your question concerns the meaning of the first of the two iterated integrals. Since the end result has to be a scalar, the integrand of the second of the two iterated integrals has to be a 1-form on $\gamma_2$. This suggests that the result of integrating the 2-form over each slice is a 1-form. In the present case, since the form $dz_2$ does not depends on $p \in \gamma_2$, you can give a well-defined meaning to your integral (as the value of the integral over any slice), and the result is indeed $dz_2$.
This explanation was more about integration the form $dz_1 \wedge dz_2$ over a curve $\gamma_1 \subset [z_1=0]$, but this is not what your question is about.
Nevertheless, one can still interpret $(1/z_1) dz_1 \wedge dz_2$ (mind the difference with what is written in the question; is it a typo?) as a 1-form over $\mathbb{C}^2 \setminus [z_1 = 0]$ taking value in $T^*\mathbb{C}^2$ (the specific value being $dz_2$). Let $\gamma$ be a closed curve around the plane $[z_1 = 0]$. By (a somewhat generalized) Stokes theorem, we can assume that $\gamma \subset [z_2 = 0]$. In this case, the 'intuitive' explanation I gave tells us more easily that the integral is $((2\pi i)^{-1} \oint_{\gamma} z_1^{-1}dz_1) \wedge dz_2 = dz_2$.
The book Principles of Algebraic Geometry by Griffiths and Harris discusses differential forms with value in more general vector spaces (and vector bundles). This is also applied to complex calculus in higher dimension. The book Differential Forms in Algebraic Topology by Bott and Tu discusses 'fiber integration', where one integrates a $k$-differential form over a $l$-submanifold with $l \le k$ to get a $(k-l)$-differential form with value in the appropriate space.
These references are possibly too technical for your needs, so I preferred to give an informal answer here.
