Not really an answer, but a comment about the related question:

> Can one reach every cluster in that way ?

It is rather easy to check (using [sagemath][1] or [Keller's applet][2]) that for the cyclic quiver in type $D_n$, not every cluster can be reached in that way. For example, only 13 among the 14 clusters in type $D_3$ are reached.

A conjecture for the number of reached clusters for this initial quiver in type $D_n$ can be found in this [OEIS sequence][3].

**EDIT:**

Even in type $A_4$, not all clusters are reached if the initial cluster is not the equioriented Dynkin quiver.

Here is a small sage program that generates the set of reached clusters:

    def recursive_comb(carquois):
        """
        EXAMPLES::
    
            sage: Q = ClusterQuiver(DiGraph({0:[1]}))
            sage: len(list(recursive_comb(Q)))
            5
        """
        n = carquois.n()
        seed = ClusterSeed(carquois)
        initial_cluster = seed.cluster()
    
        def voisins(s):
            c = s.cluster()
            for i in range(n):
                if c[i] in initial_cluster:
                    yield s.mutate(i, inplace=False)
    
        return RecursivelyEnumeratedSet([seed], voisins, structure='graded')

**EDIT:**

Using the same program, one can see that some cluster variables are missing already for some choices of initial quiver in type $D_4$:

    sage: Q1 = ClusterQuiver(DiGraph({'b':['c'],'c':['d','e']}))
    sage: Q2 = ClusterQuiver(DiGraph({'c':['b','d','e']}))
    sage: len(set(v for c in recursive_comb(Q1) for v in c.cluster()))
    16
    sage: len(set(v for c in recursive_comb(Q2) for v in c.cluster()))
    15

**EDIT**:

Even in type $A$, some clusters can be out of reach:

    sage: Q1 = ClusterQuiver(DiGraph({'b':['c','a'],'c':['d']}))
    sage: len(list(recursive_comb(Q1)))
    37

instead of 42, the full number of clusters.

  [1]: https://www.sagemath.org/
  [2]: https://webusers.imj-prg.fr/~bernhard.keller/quivermutation/
  [3]: https://oeis.org/A165205