Suppose $I\oplus K$ is a free module, then "$KI\subseteq K\cap I$"

I think your doubt is justified.

Here’s a very simple example (where the ring is right self-injective and the ideal $I$ has zero left annihilator in $R$).

Let $R=k$, a field, and let $I=k$.

$I$ can be regarded as the (right $k$-module) summand of $F=k\oplus k$ spanned by $(1,1)$, with complement $K$ spanned by $(0,1)$.

Then the annihilator of $I$ in $F$ with the “canonical multiplication” (which I presume means coordinate-wise) is zero.

Of course, in this example you could make a different choice of the embedding of $I$ into $F$ so that it had non-zero annihilator. But the “canonical” multiplication, and hence the annihilator, depend on the choices you make.