Why is it that we can define a function when Green's theorem is zero? 
When it says that "this shows we can define a function...". Why is this? Why do we get this from greens theorem?.
We have a previous theorem that says

If $f:Ω→C$ is a continuous function in a
convex open set $Ω⊂C$ and there exists $p∈Ω$ such that $f$ is
analytical in $Ω/{p}$, then $f$ has a primitive(antiderivative) in Ω.

Assuming that $v$ is a primitive, Is this related to it? But the previous theorem only works for convex set.
 A: In the first lines above, the author uses Green's theorem to prove that the line integral around a closed loop $\alpha$
$$
\int_\alpha \left[ - \frac{\partial u}{\partial y} dx + \frac{\partial u}{\partial x} dy \right] = 0.
$$
This follows from Green's theorem and the fact that $u$ is a harmonic function ($\Delta u = 0$.)
Because the line integral around any closed loop is zero, it follows that the line integral along an open path $\gamma$ from $p$ to any point $(x,y)$ is independent of the path taken between those points.  Thus, the functions $v$ is a function of $x$ and $y$ only, and does not depend on the curve $\gamma$ used to connect them.
Proof of statement about path-independence:  If there were two paths $\gamma_1$ and $\gamma_2$ between $p$ and $(x,y)$ for which the integrals were different, we could construct a closed loop starting & ending at $p$ consisting of $\gamma_1$ and then $\gamma_2$ in reverse.  By assumption, the contributions from $\gamma_1$ and $\gamma_2$ would not cancel, so the total integral around this loop would be non-zero.  But this contradicts the statement that the integral around any closed loop must be zero.
A: The definition of $v(x,y)$ says "where $\gamma$ is any path between $p$ and $(x,y)$."  To make sure that this definition makes sense, you need to prove that you get the same value for $v(x,y)$ no matter what path $\gamma$ you use.  Green's theorem allows you to prove that.
