The Kinematic transport theorem is a very basic theorem relating time derivatives of vectors between a non rotating frame and another one that's rotating with respect to it with a uniform angular velocity.
I was trying to prove it for the special case of $3$ dimensions, and everything seems straightforward apart from the fact that I'm getting into a difficulty with obtaining the exact form of the final expression.
Following is my attempted proof:
Let's assume without loss of generality that we have a frame $\widetilde{O}$ rotating with angular velocity $\mathbf{\Omega} = (0,0,\omega)$ about the $z$ axis of another frame $O$. Let $\bf{f}$ be a vector seen by frame $O$ and $\widetilde{\mathbf{f}}$ the same vector as seen by an observer in the $\widetilde{O}$ frame, and further assume that this observer is located at position $\widetilde{\bf{r}}$ with respect to this rotating frame, at a distance $R$ away from the origin. Note that this distance $R$ is correct for both frames since their origins coincide by assumption.
It is clear that this observer, as seen from the $O$ frame is given by the position vector $\bf{r}$ as follows:
\begin{equation}
\textbf{r} = R(\cos{\omega t}, \sin{\omega t}, 0)
\end{equation}
Hence the vector $\widetilde{\bf{f}}$ is given by:
\begin{equation}
\widetilde{\textbf{f}} = \textbf{f} – \textbf{r} = (f_x – R\cos(\omega t), f_y – R\sin(\omega t), f_z)
\end{equation}
Differentiating the above with respect to time we get:
\begin{align}
\dot{\widetilde{\textbf{f}}} &= (\dot{f_x} + \omega R\sin(\omega t), \dot{f_y} – \omega R\cos(\omega t), \dot{f_z}) \\
& = \dot{\textbf{f}} + \omega R (\sin(\omega t), -\cos(\omega t), 0) \\
& = \dot{\textbf{f}} + \omega (r_y, -r_x, 0)
\end{align}
Isolating $\dot{\mathbf{f}}$ we get:
\begin{equation}
\dot{\textbf{f}} = \dot{\widetilde{\textbf{f}}} + \omega (-r_y, r_x, 0)
\end{equation}
Now as you can see, what I'm getting is slightly different than what I'm supposed to get according to the theorem, the above in fact reads:
$$\left(\frac{d\mathbf{f}}{dt}\right)_O = \left( \frac{d\mathbf{f}}{dt}\right )_\widetilde{O} + \mathbf{\Omega} \times \mathbf{r} $$
Where I've replaced $\dot{\mathbf{f}}$ and $\dot{\widetilde{\mathbf{f}}}$ with the "abstract" notation that denotes differentiating with respect to each frame $O$ and $\widetilde{O}$, just to make it look more similar to how the theorem is usually stated.
However, what I'm supposed to get according to the theorem is in fact:
$$\left(\frac{d\mathbf{f}}{dt}\right)_O = \left( \frac{d\mathbf{f}}{dt}\right )_\widetilde{O} + \mathbf{\Omega} \times \mathbf{f} $$
Where is my error? I suspect that it may have to do with how I am interpreting the operation of differentiating the vector "in the rotating frame", for example I'm not totally sure it's correct to say that: $\dot{\widetilde{\textbf{f}}} = \left( \frac{d\mathbf{f}}{dt}\right )_\widetilde{O}$
Also it's very weird that the final expression I got depends on the position of the observer in the rotating frame, but I can't find what causes this error either.
Best Answer
As mentioned in a comment, I've noticed that my method is inadequate once I saw that putting $R=0$ will lead to no rotation at all of the supposedly rotating frame. The proper way to describe vectors in different frames is, as @nickbros123 mentioned, via expressing the same vector in different coordinate bases. Once this is done correctly, it seems that the derivation is quite simple:
Let $\mathbf{b_1},\mathbf{b_2},\mathbf{b_3}$ be any orthonormal basis of the non rotating frame $O$ and $\widetilde{\mathbf{b_1}},\widetilde{\mathbf{b_2}},\widetilde{\mathbf{b_3}}$ those of the rotating frame $\widetilde{O}$, that's rotating with angular velocity $\mathbf{\Omega} = (0,0,\omega)$. So we have that:
\begin{align} \widetilde{\mathbf{b_1}} &= \cos{\omega t}\mathbf{b_1} + \sin{\omega t}\mathbf{b_2} \\ \widetilde{\mathbf{b_2}} &= -\sin{\omega t}\mathbf{b_1} + \cos{\omega t}\mathbf{b_2} \\ \widetilde{\mathbf{b_3}} &= \mathbf{b_3} \end{align}
Now the statement that for any vector $\mathbf{f}$ it must hold that $\left(\mathbf{f}\right)_O = \left(\mathbf{f}\right)_{\widetilde{O}}$ is simply that the vector is the same regardless of the basis it is expressed by, that is, if it has components $(f_x,f_y,f_z)$ with respect to $O$ and components $(\tilde{f_x},\tilde{f_y},\tilde{f_z})$ with respect to $\widetilde{O}$ then it must hold that:
$$ f_x\mathbf{b_1}+f_y\mathbf{b_2}+f_z\mathbf{b_3} = \tilde{f_x}\widetilde{\mathbf{b_1}}+\tilde{f_y}\widetilde{\mathbf{b_2}}+\tilde{f_z}\widetilde{\mathbf{b_3}} $$
Differentiating the left hand side simply yields $\left(\frac{d\mathbf{f}}{dt}\right)_O = (\dot{f_x},\dot{f_y},\dot{f_z})$ and there are no additional terms in that basis simply because by assumption $\dot{\mathbf{b_i}} = 0$ for $i=1,2,3$. Differentiating also the right hand side, we get:
\begin{align} \left(\frac{d\mathbf{f}}{dt}\right)_O &= \dot{\tilde{f_x}}\widetilde{\mathbf{b_1}}+\dot{\tilde{f_y}}\widetilde{\mathbf{b_2}}+\dot{\tilde{f_z}}\widetilde{\mathbf{b_3}}+\tilde{f_x}\dot{\widetilde{\mathbf{b_1}}}+\tilde{f_y}\dot{\widetilde{\mathbf{b_2}}}+\tilde{f_z}\dot{\widetilde{\mathbf{b_3}}} \\ &= \left(\frac{d\mathbf{f}}{dt}\right)_\widetilde{O} + \tilde{f_x}\dot{\widetilde{\mathbf{b_1}}},+\tilde{f_y}\dot{\widetilde{\mathbf{b_2}}}+\tilde{f_z}\dot{\widetilde{\mathbf{b_3}}} \end{align}
Now since we note that:
$$\dot{\widetilde{\mathbf{b_1}}} = -\omega\sin{\omega t}\mathbf{b_1}+\omega\cos{\omega t}\mathbf{b_2} = \omega\widetilde{\mathbf{b_2}} \\ \dot{\widetilde{\mathbf{b_2}}} = -\omega\cos{\omega t}\mathbf{b_1}-\omega\sin{\omega t}\mathbf{b_2} = -\omega\widetilde{\mathbf{b_1}} \\ \dot{\widetilde{\mathbf{b_3}}} = 0$$
We obtain:
\begin{align} \left(\frac{d\mathbf{f}}{dt}\right)_O &= \left(\frac{d\mathbf{f}}{dt}\right)_\widetilde{O} + \tilde{f_x}\dot{\widetilde{\mathbf{b_1}}},+\tilde{f_y}\dot{\widetilde{\mathbf{b_2}}}+\tilde{f_z}\dot{\widetilde{\mathbf{b_3}}} \\ &= \left(\frac{d\mathbf{f}}{dt}\right)_\widetilde{O} + \omega\tilde{f_x}\widetilde{\mathbf{b_2}} -\omega\tilde{f_y}\widetilde{\mathbf{b_1}} \\ &= \left(\frac{d\mathbf{f}}{dt}\right)_\widetilde{O} + \left(\mathbf{\Omega} \times \mathbf{f}\right)_{\widetilde{O}} \end{align}
Which appears to be the correct result.
Additional observations and useful resources
Despite the fact that the above argument assume a very special looking form for $\mathbf{\Omega}$ it can in fact be easily generalized, by considering that the orthonormal basis which is our starting point $\mathbf{b_1},\mathbf{b_2},\mathbf{b_3}$ can be related to any other fixed orthonormal basis by a fixed rotation. With respect to this other basis then, $\mathbf{\Omega}$ will no longer have this special form, and yet such a fixed rotation clearly leaves the rotated basis vectors as time independent, and hence does not affect the rest of the derivation.
Since writing this, I have found two very good posts that are very much related and worth reading. The first one is a derivation of the centrifugal and Coriolis force terms that appear in a rotating reference frame, which basically derives the same theorem (without naming it, which is why it took me a long time to find!) in a slightly different way. The second one is a truly beautiful mathematical treatment that much more generally derives the existence of all the known fictitious forces that arise in a non inertial frame of reference, the kinematic transport theorem can also be derived by applying the same techniques used there.