I still hope there is a complete (and easy) answer to the question, but as mentioned in my comment above, and since no one else answered so far, I give a description of the inverse map that is not completely local, but still "better" than first labelling all chains in the noncrossing partition lattice. (I just worked it out briefly, so I am not saying this is the best way of doing it, but it is easy enough to be worked out in 30 min.)
A parking function is called primitive if it is weakly increasing. Every parking function can be obtained from a primitive parking function by a series of transpositions of consecutive numbers. E.g.
$$112355 \mapsto 121355 \mapsto 211355 \mapsto 213155 \mapsto 213515 \mapsto 213551.$$
We do not yet have seen here a definition of noncrossing partitions and maximal chains in the noncrossing partition lattice. So observe that a chain in the noncrossing partition lattice can be seen as factorizations of the long cycle $(1,2,\ldots,n) \in \mathcal{S}_n$ into a product of $n$ transpositions $(ij)$. The map from these factorizations to parking functions mentioned above is then the map $\phi$ sending $c = (i_1j_1) \ldots (i_nj_n)$ to the sequence $(i_1,\ldots,i_n)$. In Stanley's article, it is shown that $\phi(factorization)$ is indeed a parking function, and that this map is a bijection. To obtain $\psi := \phi^{-1}$ we must therefore start with a sequence $(i_1,\ldots,i_n)$ and turn it into a factorization $c = (i_1j_1) \ldots (i_nj_n)$.
We do this in two steps: first, we define $\psi$ for primitive parking functions, and then obtain $\psi$ for general parking functions by a sequence of transpositions of factors $(i_kj_k)(i_{k+1}j_{k+1}) \mapsto (i_{k+1}j_{k+1})(i_k\tilde j_k)$, such that
$$(i_kj_k)(i_{k+1}j_{k+1}) = (i_{k+1}j_{k+1})(i_k\tilde j_k).$$
Observe that this uniquely determines $\tilde j_k$, and that this is not a valid procedure for any two transpositions. Nonetheless, this works for sequences coming from parking functions.
Let $seq = (i_1,\ldots,i_n)$ be a primitive parking function. $\psi(seq)$ is then given by sending the last $1$ in $seq$ to the transposition $(12)$. Afterwards, send the last $i \in 1,2$ to $(i,3)$, then the last $i \in 1,2,3$ to $(i,4)$ and so on.
For example, replacing letters in $112355$ in this way gives
$$
\begin{align*}
112355 &\mapsto 1\ (12)\ 2355 \mapsto 1\ (12)(23)\ 355 \mapsto 1\ (12)(23)(34)\ 55 \newline
&\mapsto (15)(12)(23)(34)\ 55 \mapsto (15)(12)(23)(34)\ 5\ (56) \newline
&\mapsto (15)(12)(23)(34)(57)(56)
\end{align*}
$$
It is straightforward to check that this yields indeed a factorization of the long cycle, and by construction, we have $\phi(\ (15)(12)(23)(34)(57)(56)\ ) = 112355$, as desired.
Now, to obtain the parking function $211553$, we first interchange positions $2$ and $3$, and interchange the factors such that $(i_2j_2)(i_3j_3) = (i_3j_3)(i_2\tilde j_2)$ so we obtain
$$ \psi( 121355 ) = (15)(23)(13)(34)(57)(56).$$
Here, the third factor became the second, and the second factor $(12)$ turns into $(13)$. We keep going with this procedure and obtain
$$
\begin{align*}
\psi(112355) &= (15)(12)(23)(34)(57)(56) \newline
\psi(121355) &= (15)(23)(13)(34)(57)(56) \newline
\psi(211355) &= (23)(15)(13)(34)(57)(56) \newline
\psi(213155) &= (23)(15)(34)(14)(57)(56) \newline
\psi(213515) &= (23)(15)(34)(57)(14)(56) \newline
\psi(213551) &= (23)(15)(34)(57)(56)(14).
\end{align*}
$$
This completes the construction. I didn't present that proof that both steps in the procedure actually work, but they should in fact do.
If someone wants sees a direct way of computing $\psi(213551)$, please post it! And if I should clarify anything, or if someone finds mistakes, please let me know (I haven't yet gone though all details to prove that this procedure works, and I would only do it if this is not yet known, and people are interested).
Best, Christian
