Complex coordinates in CFT The Setup: Let's say we want to study a Euclidean $\mathrm{CFT}_2$ on $\mathbb R^2$ with coordinates $\sigma^1$ and $\sigma^2$ and metric 
$ds^2 = (d\sigma^1)^2+(d\sigma^2)^2$.  
It seems to me that in the usual discussion (e.g. di Francesco, Ginsparg, Polchinski), one proceeds to consider an analytic continuation of the CFT to $\mathbb C^2$ with coordinates $z^1, z^2$ and complex metric
$ds^2 = (dz^1)^2+(dz^2)^2$
and then, one performs the coordinate transformation $z = z^1+iz^2$ and $\bar z = z^1-iz^2$.  In this way the coordinates $z$ and $\bar z$ can be considered "independent" because they are coordinates on a complex two-dimensional manifold.  Also, in these coordinates the metric becomes
$ds^2 = dz\,d\bar z$
and it becomes clear that conformal mappings consist of mappings: $(z, \bar z)\to (f(z), g(\bar z))$.
My confusion is this: Since our original theory was on $\mathbb R^2$, books say that when we do calculations, we should consider the physical theory as living on the copy of $\mathbb R^2$ embedded in $\mathbb C^2$ given by the condition $\bar z = z^*$.  But consider the mapping $(z, \bar z)\to (z^2, \bar z)$.  This is a conformal mapping on $\mathbb C^2$, but it does not map the surface $\bar z = z^*$ to itself; for example the point $(z, \bar z)=(2,2)$ gets mapped to the point $(z^2, \bar z) =(4,2)$ and $2$ is clearly not equal to $4^*$.  In particular, it seems to me that analytic continuation to a CFT on $\mathbb C^2$ enlarges the set of mappings one can have, so what relevance does it really have to the original CFT on $\mathbb R^2$?
 A: The definition of a conformal mapping in this situation is one that takes $(z,\overline{z})\to (f(z),f^*(z))$, where $f(z)$ is holomorphic. So the example you gave isn't actually conformal.
To be concrete, let's take a free boson. A conformal transformation acts as
\begin{align*}
\delta \phi&=-\epsilon v\partial \phi-\epsilon v^*\overline{\partial }\phi\\
&\approx \phi(z,\overline{z})-\phi(z+\epsilon v,\overline{z}+\epsilon v^*),
\end{align*}
where $v$ is holomorphic. So its clear that the conformal transformation acts on $z$ and $\overline{z}$ as $(z,\overline{z})\mapsto (f(z),f^*(z))$.
The confusing thing, which I think you're referring to, is that any vector $v^a$ on $\mathbf{C}^2$ such that $v^z$ is holomorphic and $v^{\overline{z}}$ is antiholomorphic satisfies $\mathcal{L}_v \delta_{ab}\propto \delta_{ab}$, and is therefore a conformal transformation. However, we know that $v^z$ and $v^{\overline{z}}$ are complex conjugates since the theory is really living on $\mathbf{R}^2$, so we should only consider such $v$'s for CFTs.
A: I think you took the wrong steps. This is my take: we first identify $\mathbb{R}^2 \cong \mathbb{C}$ the natural way. Then here we take leap to $\mathbb{C}^2$ just for computational convenience. This is because in the world of complex functions sometimes it is easier to talk not of $z \in \mathbb{C}$ alone but of $z$ and $\bar z$ $\in \mathbb{C}$ at the same time as independent variables. This because if we write $z=x+iy$, we can have functions such as $f_1=x^2+y^2=z\bar z$ and also functions such as $f_2=x^2-y^2+i2xy = z^2$. So as you can see some functions over $\mathbb{C}$ can be written as functions of $z$ alone (holomorphic functions), others as functions  of $(z,\bar z)$, and others with $\bar z$ as their only argument. So if we treat $z$ and $\bar z$ as different arguments, we naturally end up on $\mathbb{C}^2$.
About your example: You are still working with a 2-dim'l CFT. The map you defined is not a holomorphic map over $\mathbb{C}$ so by now you have strayed off the path of transformations of $\mathbb{C}$ ( or $\mathbb{S}^1$. if you consider global transformations).
