Let U be
$U=M=\mathbb{R^2} \backslash \{(x,0):x \leq 0\} $
and $x:U \rightarrow M$ be the parametrization by cartesian coordinates $(x,y)$.
Let also V be $V=\{(r,\varphi) \in \mathbb{R^2}: r >0,\varphi \in (-\pi, \pi)\}. $
Now I've got to show with the help of polar coordinates, that a coordinate transformation $V \rightarrow U$ is defined. Furthermore, I got to determine the metric coefficients of both parametrizations of M. I really don't understand why there is a coordinate transformation! I don't get it.
I will change the name of your map for obvious reasons. Consider the map $g: U \to M$ defined by $g(u) = (x(u),y(u))$. You know that $x(u) = r \cos \theta, y(u) = r \sin \theta$, where $r = \sqrt{x(u)^2+y(u)^2}$. Here $\theta \ (= \varphi)$ is just the possible angle that the vector extending from the origin to $x(u)$ can make. Well, you've thrown out the negative $x$-axis, which corresponds to all points $P \in \mathbb{R}^2$ such that the angle between $\vec{OP}$ and the positive x-axis is $\pm \pi$ i.e $\theta \in (-\pi, \pi)$.