You forgot one term when you derived $\left(\vec{v_{B/ A}}\right){\omega_0}$ with respect to the inertial reference frame. The correct would be: $$\frac{d}{dt}\left(\vec{v{B/ A}}\right){\omega_0}=\left(\vec{a{B/A}}\right){\omega_0}+\omega{/0}\times\left(\vec{v_{B/A}}\right){\omega_0}$$ in same way as you did for deriving $\vec{r{B/A}}$.
Adding something from the comments:
I will call $\mathbf{i},\mathbf{j}$ the vectors of the intertial reference frame and $\hat{\mathbf{i}},\hat{\mathbf{j}}$ the ones of the moving one
(and I'm going to be a bit loose with the notation)
Basically, for every vector you can imagine you have that: $$\mathbf{r}B = \mathbf{r}_A+\left(\mathbf{r}_B-\mathbf{r}_A\right)$$ or using your notation: $$\mathbf{r}_B = \mathbf{r}_A+\mathbf{r}{B/A}$$
In principle you could derive $\mathbf{r}A$ and $\mathbf{r}{B/A}$ in the same way, but probably $\mathbf{r}_{B/A}$ can be written much more easy in an basis moving with A, i.e. $\hat{\mathbf{i}},\hat{\mathbf{j}}$.
So we have that:
$$\mathbf{r}A=r{Ax}(t)\mathbf{i}+r_{Ay}(t)\mathbf{j}$$ and its derivative $$\frac{d\mathbf{r}A}{dt}=v{Ax}(t)\mathbf{i}+v_{Ay}(t)\mathbf{j}=\mathbf{v}_{A/o}$$
However, for the case of $\mathbf{r}{B/A}$ you have that $$\mathbf{r}{B/A}=r_{B/Ax}(t)\hat{\mathbf{i}}(t)+r_{B/Ay}(t)\hat{\mathbf{j}}(t)$$ You see the difference? 804