How do we show two functions have orthogonal actions on a set or group? If we take initially $X=\{x\in \Bbb Z[\frac12]\setminus0\}$ then we can see pretty quickly that from any given starting number, the actions of the two functions on $X$ will generate the whole set:
$f(x)=x+2^{\nu_2(2x)}$ and
$g(x)=2x$
Where $2^{\nu_2(2x)}$ is double the highest power of $2$ that divides $x$.
Here I'm using the fact that both functions have well-defined unique inverses and they commute with each other.
Then when I say their actions generate the whole set I mean that from any starting number, the free abelian group on two elements, whereby each element represents the number of compositions of either f or g, brings us to every number in the set.
To prove this, see that $f$ enumerates the odd numerators and $g$ chooses the power of $2$ denominator.
One therefore might conclude the functions are in some sense orthogonal, or form a basis for something or that they enjoy some form of independence and potence over the set.
Having come to this by my own efforts I don't know where it sits in maths nor how a mathematician would normally talk of it.
A. What language is normally used to describe this?
B. What do we call the free abelian group generated by the two functions, which is the set of all transformations between pairs of elements of the set $X$?
C. How, more generally would we approach the problem of showing that the discrete actions of a pair of functions are orthogonal in the sense that these two here are, when it might not be so obvious?
 A: I try to answer A and C. (Question B. is unclear to me. For question B, see "Added" below.) Shortly, I guess what you perceive as "orthogonality" here is actually just a certain product decomposition of a set, and functions which essentially operate on different factors of that product. So to show that this would happen in other situations, one would need to show exactly that: that the underlying set decomposes as a product, and that the functions "essentially" operate on different factors of that product (meaning that on each factor, at least one of them operates as identity). 
Here, every element of the set $X = \Bbb Z[\frac{1}{2}] \setminus\lbrace 0\rbrace$ can uniquely be written as (a power of $2$) times (an odd integer), which we can express by saying that $X$ is the direct product
$$X \simeq  2^\Bbb Z  \times (2\Bbb Z +1) \qquad \qquad (*)$$
as sets (i.e. Cartesian product; actually, this is even a product of  multiplicative monoids, but that seems to be of no importance for what follows).
Now on the first factor, which more precisely we should write as $\lbrace 2^n: n \in \Bbb Z\rbrace$, and let's call it $X_1$, define the bijective function $g_0:X_1 \rightarrow X_1$ which sends $2^n$ to $2^{n+1}$; on the second factor, which more precisely we should write as $\lbrace 2n+1: n \in \Bbb Z\rbrace$, and let's call it $X_2$, define the bijective function $f_0: X_2 \rightarrow X_2$ which sends $2n+1$ to $2(n+1)+1 = 2n+3$.
Notice that quite obviously, for two elements $2^m$, $2^n$ of $X_1$, we have $g_0^{n-m}(2^m) = 2^n$, whereas for two elements $2m+1$, $2n+1$ of $X_2$, we have $f_0^{n-m}(2m+1) = 2n+1$.
Now, with the direct product decomposition $(*)$, your $f$ is $id_{X_1} \times f_0$, and your $g$ is $g_0 \times id_{X_2}$. I.e. the functions which operate as the above defined functions on one of the respective factors, but as identity on the respective other factor.
This explains naturally why $f$ and $g$ commute, and why for two elements $x = 2^{m_1} \cdot (2m_2+1)$ and $y = 2^{n_1} \cdot (2n_2+1)$ of $X$ , we have $g^{n_1 -m_1} \circ f^{n_2-m_2}(x) = y$.

Added: As regards question B, after some discussion and clarification in the comments: The group generated by the two bijections $f$ and $g$ is called, well, the group generated by those two bijections (understood: as subgroup of the group of all bijections of $X$). Let's call it $H$. It happens to be (isomorphic to) "the" free abelian group on two generators (which, further, is isomorphic to the additive group $\Bbb Z^2$.) What is worth noting, and easy to prove from the above, is that the action of $H$ on $X$ is simply transitive, which implies (by the orbit-stabilizer theorem) that for an arbitrary but fixed $x \in X$, the map $H \rightarrow X$ given by $h \mapsto h(x)$ is bijective. A terminology sometimes used for this situation is that $X$ is a torsor over $H$. After typing this, I see it all was already mentioned by Eric Wofsey here.
