Why is $h(z)$ tending to $0$ $?$ To  prove  the  uniqueness  of  Laurent's Decomposition :
$f(z)$  is  analytic  in  the  region  $\rho\lt |z-z_0|\lt \delta $ .It  is  decomposed  as  $$f(z)=f_0(z)+f_1(z)$$  where  $f_0(z)$  is  analytic  for $|z-z_0|\lt \delta $  and $f_1(z)$   is  analytic  for  $|z-z_0|\gt \rho$ .
Let  $f$  has  another  decomposition  $$f(z)=g_0(z)+g_1(z)$$
 Then  $$g_0-f_0=f_1-g_1$$ .  Let  $h(z)$  be  defined  such  that  $$h=g_0-f_0 \ \ on\ \ |z-z_0|\lt \delta \\ h=g_1-f_1 \ \ on\ \ |z-z_0|\gt \rho $$ The  two  definitions overlap  on  the  domain $\rho\lt |z-z_0|\lt \delta $ . So  $h$  is  entire on  the  whole  of  $\mathbb C$ . 
Now  what  needs  to  be  shown  is  that  $h$  is  bounded and  tends  to  $0$  as  $z$  tends to  $\infty$ . Then by Liouville's  theorem  $h(z)=0$  all  over  $\mathbb C$  and  the  uniqueness  is  proved. That  is  the  point  I'm  stuck  at . How  to  show  $h(z)\rightarrow 0$  when  $z\rightarrow \infty$ .$?$

The theorem you are citing has another condition:

We normalize the decomposition so that $f_1(\infty) = 0$

Also, the proof of the claim says

And $h(z)$ tends to $0$ as $z\to\infty$.

So, and I don't mean this in a mean way, but all your problems originated from sloppy reading. A very common and fatal flaw in mathematics.
