In the proof of the theorem below (Springer, Linear Algebraic Groups), $T$ is a maximal torus of $G$, with dimension $1$, $B$ is a Borel subgroup of $G$ containing $T$, and $U$ is the set of unipotent elements of $B$. We fix isomorphisms $t: k^{\ast} \rightarrow T$ and $u: k \rightarrow U$. One can show that there is a nontrivial character $\alpha$ of $T$ such that $z u(a)z^{-1} = u(\alpha(z)a)$ for all $z \in T, a \in k$.
We fix an $n \in N_G(T) \setminus T$ such that $nzn^{-1} = z^{-1}$ for all $z \in T$. The complement of $B$ in $G$ is $UnB$. One can show that $$U \times B \rightarrow UnB, (y,b) \mapsto ynb$$ is an isomorphism of varieties. Since $B$ is isomorphic $T \times U$, it follows that we have an isomorphism of varieties $$\phi: k \times k^{\ast} \times k \rightarrow UnB$$ $$(x,y,z) \mapsto u(a)n t(y) u(z)$$
I don't get where these rational functions $f, g, h$ are coming from. I assume they are members of the coordinate ring of $k^{\ast}$ and can be written as Laurent polynomial. I am guessing that we are looking at the restriction of the morphism $k \rightarrow G, a \mapsto n u(a) n^{-1}$ to $k^{\ast}$, whose image is contained in $UnB$. We then interpret this morphism $k^{\ast} \rightarrow UnB$ as a map into $k \times k^{\ast} \times k$.
I have this weird thing where I stare at something for two hours, rage quit, then I get on stackexchange and type up the question. Then right after I post the question I realize how to do it.
The functions $f, g, h$ are just the compositions of $k^{\ast} \rightarrow k \times k^{\ast} \times k$ with the projection maps.