Let me expand on the answer I gave in meta. In my mind the appropriate "categorification" begins with the observation that "cups" is a unit, and in the first approach you endow only the numerator with units while in the second approach you endow both the numerator and denominator with units. This is formalized as follows. Quantities with units which take values in $\mathbb{Q}$ are torsors for the nonzero rationals $G$ under multiplication, e.g. $G$-sets $X$ which are free and transitive. The choice of unit corresponds to a choice of element of $X$ against which all other elements are measured.
In the first approach, you give a group element $g = \frac{2}{3}$ and a set element $x = \frac{3}{4} \text{ cup}$ (where $\text{cup}$ is the distinguished element of $X$) and are asked to find the unique $y \in X$ such that $gy = x$. In the second approach, you give two set elements $x = \frac{3}{4} \text{ cup}, y = \frac{2}{3} \text{ cup}$ and are asked to find the unique $g \in G$ such that $gy = x$. So I think the key here is that the group action, as a function $G \times X \to X$, treats $G$ and $X$ nearly the same, but not canonically so. (I haven't thought this through, but it's also significant that in the first problem one can just write $y = g^{-1} x$, whereas in the second problem it is actually necessary, or at least natural, to express $x$ and $y$ in terms of $\text{cup}$ to determine $g$.)
Let me expand on the answer I gave in meta. In my mind the appropriate "categorification" begins with the observation that "cups" is a unit, and in the first approach you endow only the numerator with units while in the second approach you endow both the numerator and denominator with units. This is formalized as follows. Quantities with units which take values in $\mathbb{Q}$ are torsors for the nonzero rationals $G$ under multiplication, e.g. $G$-sets $X$ which are free and transitive. The choice of unit corresponds to a choice of element of $X$ against which all other elements are measured.
In the first approach, you give a group element $g = \frac{2}{3}$ and a set element $x = \frac{3}{4} \text{ cup}$ (where $\text{cup}$ is the distinguished element of $X$) and are asked to find the unique $y \in X$ such that $gy = x$. In the second approach, you give two set elements $x = \frac{3}{4} \text{ cup}, y = \frac{2}{3} \text{ cup}$ and are asked to find the unique $g \in G$ such that $gy = x$. So I think the key here is that the group action, as a function $G \times X \to X$, treats $G$ and $X$ nearly the same, but not canonically so. (I haven't thought this through, but it's also significant that in the first problem one can just write $y = g^{-1} x$, whereas in the second problem it is actually necessary to express $x$ and $y$ in terms of $\text{cup}$ to determine $g$.)