Does the free abelian group generated by composition of $f(x)=x+ 2^{\nu_2\left(2x\right)}$ and $g(x)=2x$ define a 2-dimensional affine space on $\Bbb Z\left[\frac12\right]\setminus0$?
$2^{\nu_2(x)}$ is the highest power of $2$ that divides $x$ so e.g. $2^{\nu_2(24)}=8$.
First up, $f(2x)=2f(x)$ so these functions commute and the group generated by their composition is free abelian.
Take any $i\in\Bbb Z\left[\frac12\right]\setminus0$
Then there will exist some $n\in\Bbb Z$ such that $\lvert g^n(i)\rvert_2=1$ so $g^n(i)$ will be an odd integer.
Then for this odd integer $s=g^n(i)$, composition with $f$ will generate $O=\{\ldots s-2,s,s+2,s+4\ldots,\}$ - i.e. all odd integers.
Then given this set, $g^m(O)$ will generate all of $\Bbb Z\left[\frac12\right]\setminus0$
Therefore for any pair of elements $x,y\in\Bbb Z\left[\frac12\right]\setminus0$ there is a pair of integers $m,p$ such that $g^mf^p(x)=y$
Is this an affine space? We can't identify any given element of $\Bbb Z\left[\frac12\right]\setminus0$, but we can identify translations - although they are parametrised only by pairs of integers.
I'm trying to understand the best language in which to talk about this type of structure.
"Affine space" is not applicable here. Normally one speaks of an affine space over a field, and it is not clear what field or ring you have in mind, and in any case you have not said anything about any sort of scalar multiplication.
The right terminology here is that the set $X=\mathbb{Z}[1/2]\setminus\{0\}$ is a torsor over the group $G\cong\mathbb{Z}^2$ generated by $f$ and $g$. A torsor over a group $G$ is a set $X$ with an action of $G$ such that for any $x,y\in X$, there exists a unique $g\in G$ such that $g\cdot x=y$. If you fix any element $x\in X$, this gives a bijection between $X$ and $G$, by sending $y$ to this unique $g$.
(This is of course conceptually similar to an affine space, but not the same thing. An affine space is essentially the same thing as a torsor over the additive group of some vector space.)