correspondence between meromorphic functions on complex torus vs. $\mathbb{C}$ itself I have an integration to do along the path which is the border of a parallelogram, 
namely, I need to show using integration the sum of the orders on the complex torus is $0$. 

I reparameterized it correctly, and solved the problem, by showing that $\gamma_1,\gamma_3$ cancel and so do the other two when orientation is chosen correctly. Letting $\omega_1,\omega_2$ be my linearly independent elements of $\mathbb{C}$, I need to show that I can shift my problem over to $\mathbb{C}$ as we have not learned integration on Riemann Surfaces yet, then bring it back to the torus. In doing so I concluded that there's a 1-1 correspondence between meromorphic functions defined on the complex torus and meromorphic defined on $\mathbb{C}$ itself. 

Let 

$f: \mathbb{C} / L \rightarrow \mathbb{C}$

be meromorphic, this corresponds to 

$F: \mathbb{C} \rightarrow \mathbb{C}$ 

which is meromorphic and doubly periodic, that is $F(\omega_i+z)=F(\omega)$ for $i=1,2$ 

and is defined via $F(z)=f(z)$ for when $z \in \mathbb{C} / L$, but what do I defined $F(z)$ to be for when $z \in L$?

Then I run the integration on $F$ instead of $f$ which is the same exact work shown, then I conclude by saying that 

$0=\sum_{z \in \mathbb{C}}$ord$_z(F)$ = $\sum_{p \in X}$ord$_p(f)$

Any help or hint on my correspondence between $F$ and $f$? Since if $z \in L$, we are $0$ mod $L$, do you send $F(z)$ to $0$? I wanna say if $z \in L$ we just map $F(z)$ to 0.
Your question is unclear. With $h$ meromorphic $1,t$ periodic with no zeros/poles on $a+n+mt+\Bbb{R},a+n+mt+\Bbb{R}t$ then $$\int_{[a,a+1]\cup[a+1,a+1+t]\cup[a+1+t,a+t]\cup[a+t,a]}h(z)dz$$ $$=\int_{[a,a+1]}-\int_{[a,a+1]}+\int_{[a,a+t]}-\int_{[a,a+t]}h(z)dz=0$$
When doing so with $h(z)=f'(z)/f(z)$  we get that $f$  meromorphic $1,t$ periodic 
 has the same number of zeros/poles on the fundamental paralellogram.
Your question is unclear. With $h$ meromorphic $1,t$ periodic with no zeros/poles on $a+n+mt+\Bbb{R},a+n+mt+\Bbb{R}t$ then $$\int_{[a,a+1]\cup[a+1,a+1+t]\cup[a+1+t,a+t]\cup[a+t,a]}h(z)dz$$ $$=\int_{[a,a+1]}-\int_{[a,a+1]}+\int_{[a,a+t]}-\int_{[a,a+t]}h(z)dz=0$$
When doing so with $h(z)=f'(z)/f(z)$  we get that $f$  meromorphic $1,t$ periodic 
 has the same number of zeros/poles on the fundamental paralellogram.