It seems this notion causes some confusion not only on me.
Let me give my own definition, by taking into account with yours.

Let $X$ be an analytic manifold, and let $\pi:\tilde{X}\mapsto X$ be some universal covering of $X$. A multivalued holomorphic function on $X$ is a holomorphic function on $U$, where $U$ is some contractible open subset of $X$, if the function $\pi\circ f$ on some component $\mathcal{C}$ of $\pi^{-1}(U)$ can be analytically extended to $\tilde{X}$. Denote this function by $\tilde{f} _\mathcal{C}$

Two multivalued holomorphic functions $(f,U)$ and $(g,V)$ are equivalent, if there exists some component $\mathcal{C}$ of $\pi^{-1}(U)$ and $\mathcal{D}$ of $\pi^{-1}(V)$, such that $\tilde{f}_\mathcal{C} = \tilde{g}_\mathcal{D}$.

This definition doesn't depend on the choice of universal covering.

**Edit:**

Here is a more rigorous definition：

Definition. A multivalued holomorphic function on $X$ based at $x$ is a function $f$ in $\mathcal{O}_x$, such that one component of $\pi^* f$ can be analytically extended to $\tilde{X}$, where $\pi: \tilde{X}\mapsto X$ is some universal covering.

Denote the ring of multivalued holomorphic functions on $X$ based at $x$ as $\tilde{\mathcal{O}} _x$, which is a subring of $\mathcal{O}_x$.
Any loop based at $x$ gives an action on $\tilde{\mathcal{O}} _x$, which is essentially the monodromy.

Claim. $\tilde{\mathcal{O}} _x$ is locally free $\mathcal{O}(X)$-module with rank the order of fundemental group.

It is essentially equivalent to the defintion that a mutilvalued fundtion is defined as function on covering space. But i feel it is more intuitive and doesen't depend on the choice of covering space.

Under this definition, I can make sense of the solution of mulivalued function of some analytic differential equation, which is actually my original motivation to ask this question.