I will assume that you know how to handle smooth differential forms on smooth real manifolds, this is a necessary prerequisite for understanding complex and holomorphic differential forms on complex manifolds.
(I won't mention commutative diagrams because I don't know how to draw these here, because I don't know which ones would be useful and because I'm confident that you'll come up with the ones you find useful yourself after you have understood the definition in the most simple setting.)
Suggestion: Let's look at the most simple situation, $\mathbb{C}$ as both a one dimensional complex manifold and as a two dimensional smooth real manifold $\mathbb{R}^2$.
On $\mathbb{R}^2$ we have a global chart, the cartesian coordinates, with coordinates denoted by $(x, y)$. In every point we have a basis of the tangential space $(\partial_x, \partial_y)$ and the dual basis of differential forms in the cotangential space $(d x, d y)$. Every (smooth) differential form can therefore be written (in this chart) as
$$
d w = f_1(x, y) d x + g_1(x, y) d y
$$
where we say that $d w$ is smooth if $f$ and $g$ are.
Now we can make a coordinate change in $\mathbb{C} = \mathbb{R}^2$ to the new coordinates
$$
z = x + i y
$$
and
$$
\bar{z} = x - i y
$$
This is an example of a biholomorphic mapping of an open set $V_2 = \mathbb{C}$ to $V_1 = \mathbb{C}$, which is therefore a change of coordinates on $\mathbb{C}$ (both as a real smooth and as a complex manifold). Now we can express every differential form $w$ again in the new coordinates:
$$
d w = f_2(z, \bar{z}) d z + g_2(z, \bar{z}) d \bar{z}
$$
The $f_2, g_2$ are related to $f_1, g_1$ (how?).
Motivation for the definition of "holomorphic differential form":
The differential $d f$ of a smooth function $f$ is a real smooth differential form. Similarily, one would like to define "holomorphic differential form" in a way such that the differential $d f$ of a holomorphic function $f$ is a holomorphic differential form. Since $f$ is holomorphic iff it depends on the coordinate $z$ only, one defines a holomorphic differential form to be a differential form that does depend on the coordinate $z$ only, that is $d w$ is holomorphic iff it can be written as
$$
d w = g(z) d z
$$
with a holomorphic function $g(z)$. This definition makes explicitly use of the global coordinate chart of $\mathbb{C}$ that we defined above. So, in order to show that "holomorphic differential form" is well defined on $\mathbb{C}$ seen as a complex manifold, we need to show that coordinate changes on $\mathbb{C}$ do not map holomorphic differential forms to non-holomorphic or vice versa.
A coordinate change in our example would be any biholomorphic map of an open subset of $\mathbb{C}$ to another open subset of $\mathbb{C}$.
Suggestion: Pick an easy example, write $d w$ in real coordinates and do a coordinate transform.
After that, have another look at the two definitions you cited.
$\newcommand{\dd}{\partial}$If $\omega_{j}$ is a meromorphic $1$-form in an open set $U_{j}$ for $j = 1$, $2$, and if $\omega_{1} = \omega_{2}$ on the intersection $U_{1} \cap U_{2}$, then we have a meromorphic $1$-form on the union $U_{1} \cup U_{2}$; agreement of the representatives on the intersection guarantees well-definedness. Further, the $1$-form $\omega_{1}$ has at most one extension to $U_{1} \cup U_{2}$ by the identity theorem.
In the case of an affine plane curve, $\omega_{1} = du$ is holomorphic in $U_{1} = \{\dd f/\dd v \neq 0\}$. Since
$$
\frac{\dd f}{\dd u}\, du + \frac{\dd f}{\dd v}\, dv = 0
$$
everywhere on the curve, we have $du = -\frac{\dd f/\dd v}{\dd f/\dd u}\, dv$ in $U_{2} = \{\dd f/\dd u \neq 0\}$. If the curve is non-singular, these open sets cover, so we're in the situation of the preceding paragraph. Practically, we proceed as if $du$ is globally defined.
Added: At a point $p$ where $\dd f/\dd u(p) = 0$, the holomorphic $1$-form $(\dd f/\dd v)\, dv$ vanishes. The "culprit" turns out to be $dv$, which annihilates every tangent vector at $p$: Think of a graph having a horizontal tangent, so every tangent vector at $p$ has $v$-component zero.
Best Answer
In short, the chain rule is your friend.
In greater detail, fix a point $p$ in the domain of $\psi$. Choose two charts $\phi_1$ and $\phi_2$ in the given atlas, both containing $p$ in their domain. Using Miranda's terminology, denote by $z$, $v$ and $w$ be the local variables associated to $\phi_1$, $\phi_2$ and $\psi$, respectively. Let $f(z)dz$ be the holomorphic $1$-form with respect to the chart $\phi_1$. Since the holomorphic $1$-form with respect to the chart $\phi_1$ transforms to the holomorphic $1$-form with respect to the chart $\phi_2$ on the common domain of $\phi_1$ and $\phi_2$, the holomorphic $1$-form with respect to $\phi_2$ is given by $$(f\circ\phi_1\circ\phi_2^{-1})(v)\cdot(\phi_1\circ \phi_2^{-1})'(v)dv. \tag{1}$$
If, following Miranda, we define the holomorphic $1$-form with respect to $\psi$ by using the chart $\phi_1$, we obtain $$\big(f\circ \phi_1\circ \psi^{-1}\big)\big(w\big)\cdot\big(\phi_1\circ \psi^{-1}\big)'\big(w\big)dw.\tag{2}$$
On the other hand, if we use the chart $\phi_2$, the expression $(1)$ yields $$\big(f\circ\phi_1\circ\phi_2^{-1}\circ\phi_2\circ\psi^{-1}\big)\big(w\big)\cdot\big(\phi_1\circ\phi_2^{-1}\big)'\big(\phi_2(\psi^{-1}(w))\big)\cdot\big(\phi_2\circ\psi^{-1}\big)'\big(w\big)dw.\tag{3}$$
By $(2)$ and $(3)$, showing that the definition of the holomorphic $1$-form with respect to $\psi$ is independent of the choice of chart amounts to proving the following equality $$\big(f\circ \phi_1\circ \psi^{-1}\big)\big(w\big)\cdot\big(\phi_1\circ \psi^{-1}\big)'\big(w\big)= \big(f\circ\phi_1\circ\phi_2^{-1}\circ\phi_2\circ\psi^{-1}\big)\big(w\big)\cdot\big(\phi_1\circ\phi_2^{-1}\big)'\big(\phi_2(\psi^{-1}(w))\big)\cdot\big(\phi_2\circ\psi^{-1}\big)'\big(w\big).$$
Now, the subsequent equality clearly holds $$\big(f\circ \phi_1\circ \psi^{-1}\big)\big(w\big)= \big(f\circ\phi_1\circ\phi_2^{-1}\circ\phi_2\circ\psi^{-1}\big)\big(w\big).$$ By employing the chain rule, we additionally see that $$\big(\phi_1\circ \psi^{-1}\big)'\big(w\big)= \big(\phi_1\circ\phi_2^{-1}\circ\phi_2\circ \psi^{-1}\big)'\big(w\big)= \big(\phi_1\circ\phi_2^{-1}\big)'\big(\phi_2(\psi^{-1}(w))\big)\cdot\big(\phi_2\circ\psi^{-1}\big)'\big(w\big).$$ This proves that the definition of the holomorphic $1$-form with respect to $\psi$ is independent of the choice of chart.