I was reading this introduction to operads in which the associahedron operad $\mathcal{K}$ is explained. In section 3 there is an explanation of a truncated action of this operad. Literally it says
being homotopy associative is the same thing as saying there is an action of just $\mathcal{K}(i)$ for $i ≤ 3$.
where homotopy associative means that there is a homotopy between the operations $\mu(\mu(-,-))$ and $\mu(-,\mu(-,-))$, where $\mu$ is a product.
However, if one has an action of the whole operad $\mathcal{K}$ then in particular there is a homotopy between both operations since any two bracketing of a product are homotopic (see the picture of $\mathcal{K}(4)$ in page 2).
I am having troubles to geometrically understand what is the difference between having an action of only $\mathcal{K}(i)$ ($i\leq 3$) and having an action of the whole operad. In the first case I believe that all the homotopies between any two bracketings of a product are the same, while on the second case there is a parametrized family of homotopies.
Q: Is my interpretation correct? Can someone show me an example of this difference?