Reduction of graph chromatic number to hypergraph 2-colorability
As the other answer points out, the reduction in the original paper seems to have a bug: $H$ will not be two-colorable unless $G$ is bipartite. I couldn't quite see how to prove the reduction in the other answer works, but here is one that I think does.

The vertices of $H$ will be $\{x_{i, v}: i \in [k], v \in V(G)\} \cup \{z\}$. $H$ has an edge $f_v = \{x_{1,v}, \ldots, x_{k,v}\}$ for every $v \in V(G)$,  and, for each $(u,v) \in E(G)$, and each $i \in [k]$, an edge $\{z, x_{i,u}, x_{i,v}\}$.

Given a coloring $\chi_G:V(G) \to [k]$ of $G$ you can construct a $2$-coloring $\chi_H:V(H) \to \{0,1\}$ of $H$ by setting 

$$
\chi_H(z) = 1,\\
\chi_H(x_{i,v}) = 1 \iff \chi_G(v) = i.
$$

It's easy to check that no edge of $H$ is monochromatic.

In the other direction let $\chi_H$ be a proper 2-coloring of $H$, and assume that $\chi_H(z) = 1$. (If this is not the case just rename the colors.) Since no edge $f_v$ is monochromatic, for every $v \in V(G)$ there is some $i$ such that $\chi_H(x_{i,v}) = 1$. Pick an arbitrary such $i$ and set $\chi_G(v) = i$. This is a proper coloring, or otherwise there would be an $i \in [k]$ and an edge $(u,v) \in E(G)$ such that $\chi_H(x_{i,u}) = \chi_H(x_{i,v}) = 1$, which would make the edge $\{z, x_{i,u}, x_{i,v}\}$ of $H$ monochromatic.