The quiver given in the question has five simple modules, six which correspond to a single arrow, and the remaining representations have support as follows:
123, 124, 125, 235, 345, 1235, 12235 (note the dimension over vertex 2 is 2 in this case), 2345, 12345 (note that the 42 arrow acts as zero in the last two).
Note that the 1235 subquiver is an acyclic subquiver of type $D_4$, so it has the 12 representations which Gabriel's theorem guarantees.
I think it is clear that there is an indecomposable representation corresponding to each of the dimension vectors I listed (though ask me if that's not clear). It isn't obvious that I have successfully listed all of them, but we know there must be as many as the number of non-initial cluster variables for $D_5$, so that justifies that there shouldn't be any others.
Edited, following Matt's suggstion, to add that I am imposing the relations that, for each arrow, the sum of the paths from the head of the arrow to the tail of the arrow is zero. These are the relations which define the cluster tilted algebra. (Without imposing those relations, the quiver has far more indecomposable representations -- there are infinitely many and, in fact, in a precise sense, there are so many as to be unclassifiable.)
Edited again to add that it is easy to classify all the representations for any "kite"-type quiver. That is to say, any quiver like the one drawn, but with a tail of $n-4$ vertices, instead of the tail of one vertex given by the OP.
To fix numbering, let $n$ be the vertex at the tip of the kite (numbered 4); let $n-3$, $n-4$, ..., $1$ be the tail of the kite, and let $n-1$ and $n-2$ be the other two vertices (3 and 5 in the picture).
- There are $(n-1)(n-2)$ representations of the $D_{n-1}$ subquiver avoiding $n$; the remaining representations all include $n$.
- There are $n-2$ representations that are supported at $n$ and whose additional support is along the tail of the kite only.
- There are $n-3$ representations that include the whole kite and part of the tail.
- There are another 3 representations including $n$ and at least one of $n-1$ and $n-2$, but nothing else.
The total is $n(n-1)$, which is the right answer, so this must be all of them.