The color changes are due to different concentrations of supposedly $\ce{I2}$ (which causes the yellow/amber color), $\ce{I-}$ (colorless) and $\ce{I3-}$ (dark blue).
Since iodine ($\ce{I2}$) is barely soluble in water, I do not understand why in this case it would be.
If one was to put just iodine in water and mix it thoroughly, the color change of said water would not be as visible to the extent of the Briggs-Rauscher reaction.
The color is in fact caused by the elemental iodine. At the 'yellow' stage of reaction the concentration of iodide is low, so the triodide anion (yellow at low concentration and red-brown at high) does not form, hence there is no formation of the blue triodide-amylose complex.
The OP wonders how the iodine concetration can be high enough to give the solution its yellow coloration, given the low solubility of elemental iodine in water?
First let me make a remark that crystalline iodine is very sluggish to dissolve in water to reach its maximal concentration. Even when making potassium iodide - iodine solutions - it takes a lot of stirring, anyone who worked in an analytical lab knows that.
But this is not the answer to the question. You have to consider that the iodine in solution does not come from a solid phase going to the liquid phase - it is generated in a homogenous chemical reaction. Therefore, it can reach concentrations far beyond the maximal solubility, forming a metastable supersaturated solution. This stage of the oscillation does not last long enough to allow nucleation and precipitation of the elemental iodine. The subsequent increase of iodide concentration causes formation of triodide, which along with starch gives the blue color and so on.
The colour changes are due to the indicator. In the case of the Briggs-Rauscher reaction, starch is included which causes a blue colour when complexed with iodine and iodide (high concentrations of $\text{I}_3^-$. When the concentration of $\text{I}_2$ is high, the reaction mixture is yellow due to the colour of iodine.
