After a change of variables, the answer is sequence A033282 in the Encyclopedia of Integer Sequences: $T(n,k)$ is the number of diagonal dissections of a convex $n$-gon into $k+1$ regions. The page gives the wonderful formula,
$$T(n,k) = \frac{1}{k+1}\binom{n-3}{k}\binom{n+k-1}{k},$$
for relevant values of $k$.
I find it easier to also keep track of the dual Stasheff polytope, which can be realized as a simplicial complex based on dissections of a convex polygon. The polytope $K_n^*$ has a vertex for each diagonal of an $(n+1)$-gon, and it has a face for every collection of disjoint diagonals. So, in terms of your original parameters, $K_n$ has $T(n+1,n-2-k)$ faces of dimension $k$.
Also: One reason that I like the dual Stasheff polytope is that it has an amazing infinite generalization called the Hatcher-Thurston arc complex. You again take collections of disjoint arcs that connect marked points, but in the generalization you can take any surface with or without boundary, as long as it has at least one marked point total and at least one on each boundary component. (And I suppose in the disk case it needs at least three marked points.) Each isotopy class of arcs is a vertex, and each disjoint collection is a face. It is a combinatorial model of Teichmüller spaces or moduli spaces of curves (with the requisite marked points).
Gil Kalai in the comments asks for a few more details of the Hatcher-Thurston arc complex, and he gives a reference to one of the original papers, "On triangulations of surfaces, Topology Appl. 40 (1991), 189–194," by Allen Hatcher. Briefly: Suppose that $\Sigma$ is a fixed surface with some marked points. There should be enough marked points so that there exists at least one generalized triangulation of $\Sigma$ whose vertex set is the marked points. A question that could be taken as motivation is the following: Can you find a complete set of moves on triangulations, moves on moves, moves on moves on moves, etc.? Whether the set of moves is complete is not entirely a rigorous question, but there is an interesting answer. The moves and higher moves just come from erasing edges of the triangulation. The main theorem is that the resulting simplicial complex is contractible. The mapping class group acts on the complex, and it acts freely on the high-dimensional simplices.
More precisely, the arc complex has a vertex $v$ for every isotopy class of an arc between two of the marked points, among properly embedded arcs that miss all of the marked points in the interior. If a collection of arcs can be made disjoint after isotopy, then the corresponding vertices subtend a simplex. The disk case is an exception in which the arc complex is not quite contractible, but rather a sphere.
This is indeed a mystery. I presume the question refers to wedges of spheres of the same dimension, where there's a simple criterion (n-dimensional and (n-1)-connected, for some n). For wedges of spheres of different dimensions I don't know any such criterion. Even when it's known that a complex has the homotopy type of a wedge of n-spheres, it can be difficult to find cycles representing a basis for the homology. Proofs of (n-1)-connectedness are often by induction and hence not really very enlightening, in my experience with complexes arising in combinatorial low-dimensional topology. Ideally such a proof would proceed by showing that after deleting the interiors of some top-dimensional cells, the resulting subcomplex was contractible, hopefully by an explicit contraction. There's even one important case, Harvey's curve complex of a surface, where the complex has higher dimension than the wedge of spheres that it's homotopy equivalent to. It almost seems like a metatheorem in this area that any naturally-defined complex is either contractible or homotopy equivalent to a wedge of spheres. I can't think of any counterexamples, just off the top of my head. Perhaps in other areas the proofs are more enlightening.
Best Answer
I have some idea, but I'm not sure it works well.
First I give a triangulation to $CP^1$: let's say that $CP^1$ is $C$ with a point $\infty$ added. Then we can fix 6 vertices, $0,1,-1,i,-i,\infty$ and divide $CP^1$ in 6 vertices, 12 edges and 8 triangles. Each complex number belongs to one of the 8 triangles according to the sign of its real part, the sign of its imaginary part and whether its modulus is greater or lower than 1.
So there are 26 types of points in $CP^1$: 3 choices for the sign (or zero) for the imaginary part, 3 choices for the sign (or zero) for the real part, 3 choices for the sign of the logarithm of the modulus would yield 27, but it is not possible to have 0 real, 0 imaginary and 1 modulus (we say that $\infty$ has modulus greater than 1, but its imaginary and real parts are 0). Note that a complex number (not $\infty$) has only 25 types: the ones different from the $\infty$ type.
More generally, every homogeneous $n+1$-tuple $[z_0,\dots, z_n]$ in $CP^n$ has exactly one representative whose first nonzero coordinate is 1. The following coordinates are just complex numbers, so each of them is in one of the 25 types stated before. So we associate to a point in $CP^n$ the following data: 1) The position of the first nonzero coordinate, which is some integer $0\leq k\leq n+1$; 2) the $n-k$-tuple of the types of complex numbers in the preferred representative.
Points with the same data are in the same internal part of the same simplex, and the dimension of a simplex is given by the number of < and > appearing in point 2 (I have no time now to explain in detail, but I hope it's quite clear).
For example, vertices are points of the form $[0,0,0,1,i,0,-i,0,1,-1,0]$, that is, $n+1$-tuplesmade just of $0,1,-1,i,-i$ with the first nonzero coordinate equal to 1.