If the $U_i$'s are assumed to be connected (as in Tom Goodwillie's comment) the answer is yes. There are counterexamples if the $U_i$ are not connected.
The hypothesis implies that the differentials $df_j$ patch together to define a differential form $\omega$ on $D \setminus 0$ that is never 0 (on $D \setminus 0$). Integration of $\omega$ defines a covering space of $D \setminus 0$: that is, we can choose a basepoint,
say $* = 1/2$, elements of the covering space $Y$ consist of points $z \in D \setminus 0$ together with a value that can be obtained by integrating $\omega$ along some path in $D \setminus 0$
from $*$ to $z$.
If $\omega$ is the differential of an injective function on each of finitely many
connected $U_j$'s (following Tom Goodwilli's comment), then each $U_j$ can be lifted to the covering space $Y$. This
implies that the branched cover has finitely many sheets. Since $\pi_1(D \setminus 0)$ is abelian, $Y$ an $n$-fold cyclic cover isomorphic to $z \mapsto z^n$. In these coordinates,
pullback of $\omega$ to $Y$ can be integrated to give a function $g$ from $Y$ to the Riemann sphere such that satisfies $g(\zeta y) = g(\zeta) + C$ for some constant $C$, where $\zeta$ is a
primitive $n$th root of unity. But then $n*C = 0$, so $C = 0$, so $g$ comes from a function on $D \setminus 0$; since it is finite-to-one near 0, by Picard's theorem this is a removable singularity (as a map to the Riemann sphere), it extends to a meromorphic function on $D$, and its differential is therefore meromorphic.
If the $U_j$'s are not assumed connected, the covering space $Y$ still exists, but it need not have
finitely many sheets. In the infinite-sheeted case, the covering space is the universal cover of $D \setminus 0$ isomorphic to $z \mapsto \exp(z)$ from the right halfplane $Re(z) < 0$ to $D \setminus 0$. The integral of the pullback of $\omega$ to the right halfplane
is a function that has the form $g(z) = a z + g_0(z)$, where $g_0(z)$ is the pullback of a function $f_0$ defined on $D \setminus 0$. (The linear term takes care of the integral
of $\omega$ on a loop around the origin). The function $f_0$ must be an immersion (locally univalent function) from $D \setminus 0$ to $\mathbb C$. Such functions can be rather wild, for example, $f_0(z) = exp(1/z)$. The integral of $\omega$ itself could then be expressed as the multi-function $f(z) = z^a f_0(z)$, where $a$ is any complex number.
Claim for any complex number $a$ and any locally univalent function $f_0: D \setminus 0 \to \mathbb C$ there is a finite cover $U_i$ of $D \setminus 0$ (where the $U_i$ are not connected) such that on each $U_i$ $\omega$ is the differential of a univalent function $f$.
Proof of Claim: Cover $D \setminus 0$ by countably many open sets such that the integral
of $\omega$ in each set is univalent. Every cover of a 2-manifold has a refinement that is locally at most 3-to-1, so its nerve is a 2-complex. (This is one characterization of topological dimension). It has a further refinement $U$, corresponding to the barycentric subdivision, by a covering that can be partitioned into three parts $U = A \cup B \cup C$
where the elements of $A$ are disjoint, the elemetns of $B$ are disjoint, and the elements of $C$ are disjoint. We can integrate $\omega$ on each of the elements of $A$, $B$ and $C$, and then add suitable constants to make the images in each of $A$, $B$ and $C$ disjoint.
In summary: we've found three holomorphic functions $f_A$, $f_B$ and $f_C$ defined on
the union of $A$, the union of $B$ and the union of $C$ whose differentials all equal
$\omega$, but $\omega$ has an essential singularity at the origin.