Yes. Form Aff $(T)$; since $T$ is metrizable, Aff $(T)$ contains a countable dense set. Adjoin $f$ and the constant function $1$ to the countable dense set (creating a larger countable dense set), and let $G$ be the subgroup of Aff $(T)$ generated by this enlarged set. This is of course countable, and dense.

Therefore, equipped with the strict ordering ($G^+$ consisting of $0$ together with all the elements in $G$ that are bounded below away from zero on $T$), $G$ is a simple partially ordered group, and it is well known that dense subgroups of Aff of a Choquet simplex with respect to the strict ordering satisfies interpolation. Hence $G$ is a simple dimension group [using the result that a countable unperforated partially ordered group with interpolation is a dimension group, i.e., a limit of simpicial groups], and with respect to the order unit $u = 1$, $S(G,u)$ is naturally identified with $T$ (that is, the normalized traces of $G$ are the points of $T$).

It is easy to check that with $u = 1$ (the constant function), the map $(G,u) \to ({\rm Aff\ }S(G,u),1)$ given by $g \mapsto \hat g$ where $\hat g(s) = s(g)$ is equivalent to the inclusion $(G,1) \subset {\rm Aff } (T),1)$, and in such a way that $\hat f$ is sent to $f$.

More generally, we can do this for any countable family of $f$s, rather than just a single one.

For all of the unjustified statements here, see Goodearl's book on partially ordered abelian groups. The use of countable subgroups as here is a standard construction.

Yes. Form Aff $(T)$; since $T$ is metrizable, Aff $(T)$ contains a countable dense set. Adjoin $f$ and the constant function $1$ to the countable dense set (creating a larger countable dense set), and let $G$ be the subgroup of Aff $(T)$ generated by this enlarged set. This is of course countable, and dense.

Therefore, equipped with the strict ordering ($G^+$ consisting of $0$ together with all the elements in $G$ that are bounded below away from zero on $T$), $G$ is a simple partially ordered group, and it is well known that dense subgroups of Aff of a Choquet simplex with respect to the strict ordering satisfy interpolation. Hence $G$ is a simple dimension group [using the result that a countable unperforated partially ordered group with interpolation is a dimension group, i.e., a limit of simpicial groups], and with respect to the order unit $u = 1$, $S(G,u)$ is naturally identified with $T$ (that is, the normalized traces of $G$ are the points of $T$).

It is easy to check that with $u = 1$ (the constant function), the map $(G,u) \to ({\rm Aff\ }S(G,u),1)$ given by $g \mapsto \hat g$ where $\hat g(s) = s(g)$ is equivalent to the inclusion $(G,1) \subset {\rm Aff } (T),1)$, and in such a way that $\hat f$ is sent to $f$.

More generally, we can do this for any countable family of $f$s, rather than just a single one.

For all of the unjustified statements here, see Goodearl's book on partially ordered abelian groups. The use of countable subgroups as here is a standard construction.