I have been thinking about this problem some more recently, and here is what I have come up with.
Let $\Omega$ be a probability space, then a random variable $X$ is a measurable function $X: \Omega \to \mathcal{X}$, where $\mathcal{X}$ is a measurable space ($\mathcal{X}$ has a designated $\sigma$-algebra, and $X$ is measurable with respect to this $\sigma$-algebra and the $\sigma$-algebra on $\Omega$). The distribution of $X$ is just the pullback measure on $\mathcal{X}$, i.e. $\mathbb{P}_{\mathcal{X}}(A) = \mathbb{P}_{\Omega}(X^{-1}(A))$. Then a statistic of $X$ is any measurable* function $f: \mathcal{X} \to \mathcal{Y}$, where $\mathcal{Y}$ is another arbitrary measurable space.
Given two statistics $f: \mathcal{X} \to \mathcal{Y}$, $g: \mathcal{X} \to \mathcal{Z}$, what does it mean for "$g$ to be a function of $f$"?
As far as I can tell, it seems to mean that there exists a measurable** function $h: \mathcal{Y} \to \mathcal{Z}$ such that $g = h \circ f$, i.e. that $g$ can be factored through by $f$.
(In other words, "$g$ must be well-defined as a function on $f(\mathcal{X}) \subseteq \mathcal{Y}$".)
So when is such factoring possible? Let's think in terms of equivalence relations. Specifically, define the equivalence relation $\sim_f$ on $\mathcal{X}$ by $x_1 \sim_f x_2 \iff f(x_1) = f(x_2)$, likewise, define the equivalence relation $\sim_g$ on $\mathcal{X}$ by $x_1 \sim_g x_2 \iff g(x_1) = g(x_2)$.
Then in order for $g$ to be factorable by $f$, the equivalence relations $\sim_f$ and $\sim_g$ need to be compatible with each other, in the sense*** that for any $x_1, x_2 \in \mathcal{X}$, $x_1 \sim_f x_2 \implies x_1 \sim_g x_2$, i.e. $g$ can't take two elements which are equivalent under $f$ and map them to values which aren't equivalent under $g$, i.e. "$g$ can't undo the information reduction previously performed by $f$".
In other words, $g$ has to be well-defined as a function on $\mathcal{X}/\sim_f \cong f(\mathcal{X})$, i.e. there exists has to exist a function $\tilde{g}: \mathcal{X}/\sim_f \to \mathcal{Z}$ such that $g = \tilde{g} \circ \pi_f$, where $\pi_f$ is the canonical projection $\mathcal{X} \to \mathcal{X}/\sim_f$. (For those uncomfortable with abstract non-sense, $\pi_f$ is essentially $f$, and $\tilde{g}$ is essentially $h$. The above formulation just makes analogies with other situations more clear.)
In simplest possible words, $g$ can be written as function of $f$ if and only if, for any $x_1, x_2 \in \mathcal{X}$, $f(x_1) = f(x_2) \implies g(x_1) = g(x_2)$.
For example, take $\mathcal{X} = \mathcal{Y} = \mathcal{Z} = \mathbb{R}$ and $X$ an arbitrary real-valued random variable, then $g: x \mapsto x^2$ can be written as a function of $f: x \mapsto x$, but not vice versa, because $x_1 = x_2 \implies x_1^2 = x_2^2$, but $1^2 = (-1)^2$ but $1 \not= -1$.
In particular, assume that every equivalence class under $\sim_f$ is a singleton (i.e. $f$ is injective). Then $g$ can always be written as a function of $f$, since $\mathcal{X}/\sim_f \cong \mathcal{X}$, i.e. $f(x_1) = f(x_2) \implies x_1 = x_2$ means that $x_1 = x_2 \iff f(x_1) = f(x_2)$ (in general, for not-necessarily injective $f$, only one direction holds), so our condition becomes $x_1 = x_2 \implies g(x_1) = g(x_2)$, which is trivially satisfied for any $g: \mathcal{X} \to \mathcal{Z}$. (To define $h$, it can do anything it wants on $\mathcal{Y} \setminus f(\mathcal{X})$ as long as it's measurable, and then for any $y \in f(\mathcal{X})$, i.e. such that $y = f(x)$ for some $x \in \mathcal{X}$, define $h$ to be $h: y = f(x) \mapsto g(x)$. This is well-defined when $f$ is injective because there is a unique $x \in \mathcal{X}$ such that $f(x) = y$. More generally, this is only defined when, regardless of which $x$ we choose in $f^{-1}(y)$, $g(x)$ still is the same value, i.e. $f(x_1)=f(x_2)\ (=y) \implies g(x_1)=g(x_2)$.)
Also, looking at Theorem 3.11 in Keener, its statement is kind of clunky, but thinking in the above terms, I believe it can be re-written as:
Suppose $T$ is a sufficient statistic****. Then a sufficient condition for $T$ to be minimal sufficient is that it can be written as a function of the likelihood ratio.
From this it becomes immediately clear that the likelihood ratio has to itself be minimal sufficient.
This also leads to the conclusion that:
If there exist $x_1, x_2 \in \mathcal{X}$ such that $f(x_1)=f(x_2)$ but $g(x_1) \not= g(x_2)$, then $g$ can not be written as a function of $f$, i.e. there exists no function $h$ with $g = h \circ f$.
Thus the condition isn't actually as difficult to show as I had thought.
*Keener doesn't address the issue of whether a statistic needs to be a measurable or just an arbitrary function or not. However, I am pretty sure that a statistic has to be a measurable function, because otherwise we couldn't define a distribution for it, i.e. a pullback measure.
**If $h$ were not measurable, we would have a contradiction because both $f$ and $g$ are measurable and the composition of measurable functions is again measurable. At the very least, $h$ has to be measurable restricted to $f(\mathcal{X}) \subseteq \mathcal{Y}$, although I think this would mean in most reasonable cases that $h$ would have to agree on $f(\mathcal{X})$ with a function that is measurable on all of $\mathcal{Y}$ (take $h|_{f(\mathcal{X})}$ on $f(\mathcal{X})$ and e.g. $z$ on $Y \setminus f(\mathcal{X})$ if there exists a measurable point $z \in \mathcal{Z}$, note that both $f(\mathcal{X})$ and $Y \setminus f(\mathcal{X})$ should be measurable in $Y$) so w.l.o.g. $h$ can be assumed to be measurable on all of $\mathcal{Y}$.
***At least this is necessary and sufficient for the existence of an arbitrary function factoring through $g$ and over $f$, and I think ** implies that if such an arbitrary function exists, this function also must be measurable, since both $f$ and $g$ are, i.e. it really would be a statistic $\mathcal{Y} \to \mathcal{Z}$.
****The condition given is equivalent to $T$ being sufficient by the factorization theorem, 3.6.
