Confusion have vector-nature of velocity in rotating frame Consider an inertial frame of reference, $s$, (henceforth coordinates in this system will be small letters), and a frame $S$ rotating coaxially ($z$-axis) with respect to this frame (all coordinates in big letters), with constant angular velocity $\omega$. 
Now consider a particle with position $\mathbf{x}$ in $s$, so that its coordinate in $S$ can be written as $\mathbf{X} = T(\theta)\mathbf{x}$, where $T(\theta)$ is the appropriate rotation/transformation matrix for angle $\theta$. (Note: Consider motion in $s$ in only plane $xy$, and $z$ axis is common, so motion in $S$ is in plane $XY$).
So $\mathbf{\dot{X}} = T(\theta) \mathbf{\dot{x}} + \omega\dfrac{dT}{d\theta}\mathbf{x} $.
But any vector in $s$ must satisfy that its coordinates will have to transform by the $T(\theta)$ transformation rule (to qualify as a vector).
By that logic, velocity in $S$, $\mathbf{V}$ must be such that $\mathbf{V} = T(\theta)\mathbf{\dot{x}}$, as $\mathbf{\dot{x}}$ is velocity is $s$. This must imply that $\mathbf{\dot{X}}$ is not the velocity in the $S$ frame, as it has an extra term as shown above.
What is going wrong here?
 A: If you write 
$$
\boldsymbol{X}=T(\theta)\boldsymbol{x}
$$
then indeed 
$$
\dot{\boldsymbol{X}}=\dot{T}(\theta)\boldsymbol{x}+T(\theta)\dot{\boldsymbol{x}}\tag{1}
$$
Thus, $\dot{\boldsymbol{X}}$ has two parts: one is due to the motion of the particle in the rotating frame - that's $\dot{\boldsymbol{x}}$ - and the other is due to the velocity of the rotating frame w/r to the inertial frame - that's the $\dot{T}(\theta)\boldsymbol{x}$ part.  
A: If the position vector of the particle with respect to $\:s\:$(1) is $\:\mathbf{x}\:$ then its velocity $\:\mathbf{V}\:$  with respect to $\:S\:$ is the resultant of two velocities : 
\begin{equation}
\mathbf{V}=[\mathbf{v}]_{S}+\mathbf{W}
\tag{01}
\end{equation}
where $\:[\mathbf{v}]_{S}\:$ is its velocity with respect to $\:s\:$ but expressed in $\:S-$coordinates and $\:\mathbf{W}\:$ its velocity due to the motion (here rotation) of system $\:s\:$ as a whole with respect to $\:S\:$.
Now
\begin{equation}
\mathbf{v}=\mathbf{\dot{x}}
\tag{02}
\end{equation}
and there is no objection, as you claim, that its expression in $\:S-$coordinates is
\begin{equation}
[\mathbf{v}]_{S}=T(\theta)\mathbf{\dot{x}}=T(\theta)\mathbf{v}
\tag{03}
\end{equation}
The extra term we have is $\:\mathbf{W}\:$
\begin{equation}
\mathbf{W}=\dfrac{\mathrm{d}T}{\mathrm{d}t}\mathbf{x}=\dot{T}\mathbf{x}
\tag{04}
\end{equation}
and if we express $\:\mathbf{x}\:$ in $\:S-$coordinates
\begin{equation}
\mathbf{x}=T^{-1}(\theta)\mathbf{X}=T^{\boldsymbol{\top}}(\theta)\mathbf{X}
\tag{05}
\end{equation}
then 
\begin{equation}
\mathbf{W}=\dot{T}T^{\boldsymbol{\top}}\mathbf{X}
\tag{06}
\end{equation}
But
\begin{equation}
TT^{\boldsymbol{\top}}=I=T^{\boldsymbol{\top}}T \quad \Longrightarrow \quad\dot{T}T^{\boldsymbol{\top}}=-\left(\dot{T}T^{\boldsymbol{\top}}\right)^{\boldsymbol{\top}}
\tag{07}
\end{equation}
that is the $\:3\times 3\:$ matrix $\:\dot{T}T^{\boldsymbol{\top}}\:$ is antisymmetric, so it could be expressed as 
\begin{equation}
   \dot{T}T^{\boldsymbol{\top}} \equiv
    \begin{bmatrix}
         0&-\Omega_3&+\Omega_2\\
          &&\\
         +\Omega_3&0&-\Omega_1\\
         &&\\
         -\Omega_2&+\Omega_1&0
       \end{bmatrix}
       \:=\:\boldsymbol{\Omega}\:\boldsymbol{\times}    
\tag{08}
\end{equation}
and finally
\begin{equation}
\mathbf{W}=\boldsymbol{\Omega}\boldsymbol{\times}\mathbf{X}
\tag{09}
\end{equation}

(1)
since we talk about kinematics and not kinetics there is no need for any one of the systems to be inertial.

