In fact, if $C = \text{rank}(I_*(S^3_0(K)))$, we have $$f(k) = \text{rank}(I_*(S^3_{1/k}(K))) = kC.$$ This holds with any coefficient field.

Floer's exact sequence gives us an exact triangle relating $I_*(S^3_0) \to I_*(S^3_{1/k}) \to I_*(S^3_{1/(k+1)}).$ As a result, we have $f(k+1) \le f(k) + C$, and $f(1) = f(0) = C$.

But there's also the exact triangle of Culler, Daemi, and Xie, displayed as (1.4) [here](https://londmathsoc.onlinelibrary.wiley.com/doi/abs/10.1112/topo.12137), relating $I_*(S^3_{1/(k-1)}) \to I_*(S^3_{1/k})^2 \to I_*(S^3_{1/(k+1)})$. From this we have $2f(k) \le f(k-1) + f(k+1)$ for $k \ge 2$ (and for $k=1$ we get $f(2) = 2f(1)$). 

Now $f(k) = kC$ can be proven by induction.

As a result, it follows that in Floer's triangle the connecting map $I_*(S^3_0) \to I_*(S^3_{1/k})$ is always zero, and in Culler-Daemi-Xie's triangle the connecting map $I_*(S^3_{1/(k+1)}) \to I_*(S^3_{1/(k-1)})$ is zero, as well.

At this point, if you know that the displayed maps in the CDX triangle above preserve the $\Bbb Z/2$-grading, you can argue by another induction using the CDX sequence that $I_*(S^3_{1/k}(K)) \cong I_*(S^3_1(K))^{\oplus k}$ as $\Bbb Z/2$-graded vector spaces.