Why not look at Stasheff's original paper? He does give a point-set model (where $K_{n+2}$ is a compact convex semialgebraic subset of $\mathbb{R}^n$) and describes explicitly the substitution maps $\text{sub}_i: K_m \times K_n \to K_{m+n-1}$ which are collectively tantamount to the operad structure.

- James Dillon Stasheff, Homotopy Associativity of H-Spaces I, Transactions of the American Mathematical Society Vol. 108, No. 2 (Aug., 1963), pp. 275-292.

There are other descriptions which realize the associahedra as linear polytopes (which I find more pleasant); the name of the late J.-L. Loday sticks prominently in my mind in this regard. See here for instance.

**Edit:** The universal property description of the associahedral operad (or Stasheff operad; I've used both terms), mentioned in my comment under the question, might not be well-known; it might be good to describe intuitively what is going on there. I should add that I have no idea where this observation might appear in the literature, but that I would also have a hard time believing that the observation is originally due to me. :-) See this $n$-Category Café post for some related background.

The statement is that the Stasheff operad is initial among non-permutative non-unital operads $M$ such that each $M_n$, for $n \geq 2$, carries a basepoint and an $I$-module map $I \times M_n \to M_n$ (for the multiplicative monoid $I = [0, 1]$) that contracts $M_n$ to that basepoint. It should be noted that the correct definition of *non-unital* operad is one that uses operations $sub_i: M_m \times M_n \to M_{m+n-1}$ for $1 \leq i \leq m$, not a less expressive definition which merely involves a structure $M \circ M \to M$ where $\circ$ is the well-known substitution product on graded spaces (see Tom Leinster's book for details on such substitution products, monoids of which being unital operads). In the Stasheff model, we could take the barycenter of each associahedron as its basepoint $p$; the contracting homotopy is given by $x \mapsto t x + (1-t)p$, taking advantage of convexity for the Stasheff model.

In the initial such structure, there is no operad identity, so $M_1$ is empty, and so is $M_0$. For $M_2$, all we know is that it has a basepoint $m_2$ (so $M_2$ is the cone on an empty space). For $M_3$, all we know is that there are points $l = sub_1(m, m)$ (think $(xy)z$) and $r = sub_2(m, m)$ (think $x(yz)$) plus a contracting homotopy to a basepoint $m_3$, so $M_3$ is a cone on the space $\{l, r\}$. This cone is of course an interval. Continuing up the inductive ladder, $M_4$ is obtained by taking the cone of a space obtained by pasting together five spaces corresponding to images $sub_1(M_2 \times M_3)$, $sub_2(M_2 \times M_3)$, $sub_1(M_3 \times M_2)$, $sub_2(M_3 \times M_2)$, $sub_3(M_3 \times M_2)$; the pasting is in accordance with generalized operadic associativity. The pasting gives the boundary of a pentagon; the cone itself is the famous Stasheff pentagon. And so on.

Of coursethere are explicit concrete descriptions, as I indicated in my answer. There is also a universal property description, where $K$ is initial among non-unital topological operads $M$ for which each $M_n$ is equipped with a basepoint and specified contracting homotopy. So if your description has this universal property, then that is good enough. $\endgroup$ – Todd Trimble♦ Sep 5 '13 at 22:37