Defining meromorphic functions and forms on a Riemann surface using a single formula in a single chart I am working out of Miranda's Algebraic Curves and Riemann Surfaces. In the book, we define differential forms for a Riemann surface to be a collection of forms satisfying a compatibility condition {$\omega_\phi\ | \ \phi:U\rightarrow V \text{ a chart in the complex structure for }X$}. We then show it suffices it to define a form over a collection of charts that cover $X$. Later on, the author writes the following:

The definition of a meromorphic or holomorphic 1-form $\omega$ suggests that in order to define $\omega$ on a Riemann surface $X$, one must give local expressions for $\omega$ (of the form $f(z)dz$) in each chart of an atlas for $X$. In fact, one can define $\omega$ by giving a single formula in a single chart. This is sufficient to determine $\omega$ by the Identity Theorem for meromorphic functions and forms: if two meromorphic 1-forms agree on an open set, they must be identical.

My difficulty is unpacking why this statement is true. I understand the statement of the Identity Theorem, but I don't see how one can apply it in this situation. Here is the line of thought that I have attempted so far.
'Define' a holomorphic 1-form $f(z)dz$ on some chart $\phi_1:U_1\rightarrow V_1$, where $p\mapsto z$. Now grab a chart with non-trivial intersection with $\phi_2$, say $\phi_2:U_2\rightarrow V_2$, where $p\mapsto w$. All we can do at this point is to extend our 1-form by defining it using the transformation formula with $T=\phi_1\circ \phi_2^{-1}$ to obtain $g(w)dw:=f(T(w))T'(w)dw$ for the chart $\phi_2$. However, if I am understanding things correctly, this extension isn't really an extension because the new form $g(w)dw$ is only defined on $\phi_2^{-1}(U_1\cap V_2)$. I imagine that this is where the Identity Theorem is supposed to be leveraged, but I am not seeing where we can apply it.
Edit:
Maybe adding some motivation for the question can clarify what I am getting at. Let $X$ be a smooth affine plane curve defined by $f(u,v)=0$. We want to show that $du$ and $dv$ define holomorphic 1-forms on $X$.
Examining $du$, let $\phi:U\rightarrow V$ be a chart where $\partial f / \partial v$ is non-zero on $U$ so that via the invertible function theorem $\phi: (u,v)\mapsto u$ is indeed a chart. Then we can define on this chart the 1-form $\omega=1du$.The transition maps can't take things outside of of the set $U$, since they are only possibly defined on intersections of $U$ with other sets - so how can one actually extend it past that?
 A: $\newcommand{\dd}{\partial}$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.
