Thanks for the answer! I apologize for answering with a delay (am on biking trip in Corsica). Yes, Tom Goodwillie is right about the necessity of the connectiveness condition. Bill Thurston's answer, however, does not convince me. I will explain why:
Call $\omega$ the holomophic form defined on the punctured unit disk. There are two cases:
The residue is zero. Then there is a holomorphic function on the unit disk such that $df=\omega.$ Thus $df=df_j$ for every j. Using the connectedness of $U_j$ one can add a constant to $f_j$ such that $f=f_j$ on $U_j$. Then one concludes easily with Picard that f is meromorphic on the unit disk.
The residu $a$ is not zero. Then integrating $\omega$ yields a cover of infinite order because each turn around the origin adds the number $a$. So here I do not agree with Bill Thurston who says the order is finite. What actually happens is that the primitive of $\omega$ is of the form: holomorphic function on the punctured disk + $a\times$ logarithm.
Now here comes my explanation of Bill Thurston's mistake when he says that the covering space is of finite order. The best way to define the Riemann R(g) surface of an analytic germ g is to say that R(g) is the connected component of g in the total space $|\mathcal{O}|$ of the sheaf of holomorphic functions. There are two natural projections defined on R(g). The first projection sends each germ to its centerpoint ("projection on the variable plane"). The second projection sends each germ to the value the germ takes in its centerpoint ("projection on the value plane"). It is the first projection that we are interested in. But when arguing that his covering space is finite above the punctured unit disk, Bill Thurston actually uses the functions $f_j$ as map, thus taking the second projection instead of the first.