Tensors time derivative in moving frames

I think this can all be dealt by being explicit with your basis vectors. So let you basis vectors be ${mathbf{e}_ileft(mathbf{r},,tright)}$, note that I allowed for time and position dependence so this is as general is it needs to be. This basis is for normal vectors, tensors, but you can define dual basis and handle the contra-variant vectors in the same way.

So that lets say we have a tensor $mathbf{T}=T^{i_1 i_2 dots i_n}left(mathbf{r}, tright) :mathbf{e}_{i_1}otimesmathbf{e}_{i_2}dotsotimesmathbf{e}_{i_n}$. Then

$frac{d}{dt}mathbf{T}=frac{d}{dt}left(T^{i_1 i_2 dots i_n}right):mathbf{e}_{i_1}otimesmathbf{e}_{i_2}dotsotimesmathbf{e}_{i_n} + T^{i_1 i_2 dots i_n}:frac{d}{dt}left(mathbf{e}_{i_1}right)otimesmathbf{e}_{i_2}dotsotimesmathbf{e}_{i_n} + dots$

I am giving this here without a proof, but I am prertty sure such proof can be established based on the universality of the tensor product.

Finally all you need is $frac{d}{dt}left(mathbf{e}_iright)$. I think it is better to evaluate this expression on case by case basis, since the general form could be ugly. For rotation, I would expect $frac{d}{dt}left(mathbf{e}_iright)=Omega^j_{:i}mathbf{e}_j$ where $Omega^j_{:i}$ is the rotation matrix (which relates directly to your cross-product, but I cannot think of a neat notation to express it without introducing metric tensors or Hodge-duals).

I am not sure if this makes things easier, but it is coordinate independent.