I will not answer probably to the question but i want to report my computation here, since maybe you can help me.
I perfectly know that the compactification is relative to a complex variety, but for a moment let's forget that and we try to do the same construction for $\mathbb{R}^2$. We consider the generating set $F=\{x,y\}$ given by coordinates and we consider the closure of the image of the function $$\theta:\mathbb{R}^2 \rightarrow S^2 \times \mathbb{P}^F, \hspace{10pt} \theta(x,y)=((x,y),[\log(|x|+2):\log(|y|+2)]).$$
If we want to close this image, i think we need to add only point to each level set of the function $$\varphi(x,y)=\frac{\log(|y|+2)}{\log(|x|+2)}$$ and two more points $\{\infty\} \times [0:1],\{\infty\} \times [1:0]$. This is equivalent to determine a suitable identification of the boundary of the disk $D^2$, as reported in this picture https://drive.google.com/file/d/0B16wiL4cSf4zbVpRcVNzaGJTYmc/view?usp=sharing.
More precisely, the points with the same color are identified on the boundary. In this way you are adding the whole space $\mathbb{P}^F \cong [0,\infty]$ and identifying the upper-right quarter of the boundary with the other ones following the identification drawn in picture. So, the space $\overline{X}^F$ should be a singular quotient of the 2-sphere $S^2$.
If we define the projection on the second factor $\pi_2:X^F \rightarrow \mathbb{P}^F$ we can consider the preimage $\pi_2^{-1}(a)$ of a point $a \in [0,\infty] \cong \mathbb{P}^F$. The picture reported below describes the topological behaviour of the preimage https://drive.google.com/file/d/0B16wiL4cSf4zRHhLTjE0Q2NnVjQ/view?usp=sharing.
Is this computation correct? Are there any mistakes in my reasoning?I met the same problems studying the MS compactification. In particular i do the same computation as above and i get something singular.
If there were no mistake in my computation, we could be able to describe at least the preimage $\pi_2^{-1}(a)$ also for the complex case. I think the result could be the following https://drive.google.com/file/d/0B16wiL4cSf4zUlJSVEZQSGYxcTg/view?usp=sharing. From this considerations, i can't see how to prove that $\overline{C^2}^F$ is homeomorphic to $S^4$.
Thanks in advance for your help.