What constitutes a proof in a system like this is a derivation which is a tree of rule applications. The above translation function is defined by (structural) recursion over that tree. Note, this is why it labels the premises in the rules with $\delta$. This is intended to be mnemonic for "derivation".
I'm guessing but $\mu$ and $\nu$ appear to need to be sequences of lambda terms, while $\phi$ is a specific lambda term (or, equivalently, a single element sequence of lambda terms). $\Gamma$ is a sequence of formulas, and $\mu$ has a lambda term for each element of that sequence. Constructs like $x,\mu$ correspond to sequence extension, extending the sequence $\mu$ with the (particular) lambda term $x$ on the left, symmetrically for $\mu,x$. The parenthesis notation, e.g. $\Gamma(C)$ and $\mu(\dots)$ correspond to (consistently) operating on an element at an arbitrary location in the sequence.
So, for example, an instance of the $\bullet R$ case would look like:
$$\left|\cfrac{\cfrac{\delta}{X,A,B,Y,Z\Rightarrow W}}{{X,A\bullet B,Y,Z\Rightarrow W}}\right|_{t_X,t_{AB},t_Y,t_Z} = \left|\cfrac{\delta}{X,A,B,Y,Z\Rightarrow W}\right|_{t_X,\pi_1(t_{AB}),\pi_2(t_{AB}),t_Y,t_Z}$$
A translation of a complete derivation would look like: $$\begin{align}
\left|\cfrac{\cfrac{A\Rightarrow A \qquad B \Rightarrow B}{A,B\Rightarrow A\bullet B}}{\cfrac{A\Rightarrow (A\bullet B)/B}{\Rightarrow ((A\bullet B)/B)/A}}\right|
& = \lambda a\left|\cfrac{\cfrac{A\Rightarrow A \qquad B \Rightarrow B}{A,B\Rightarrow A\bullet B}}{A\Rightarrow (A\bullet B)/B}\right|_a \\
& = \lambda a\lambda b\left|\cfrac{A\Rightarrow A \qquad B \Rightarrow B}{A,B\Rightarrow A\bullet B}\right|_{a,b} \\
& = \lambda a\lambda b(\left|A\Rightarrow A\right|_a, \left|B \Rightarrow B\right|_b) \\
& = \lambda a\lambda b(a, b)
\end{align}$$
