This is a problem I've struggled with before. The issue is that most textbooks are not clear enough in their definitions of the fixed and rotating reference frames. I've tried to be as comprehensive as possible in my answer, I hope it helps!
To describe the motion of a rigid body through space we must make use of two separate reference frames:
Firstly, a fixed (i.e. inertial) reference frame with a set of orthonormal basis vectors $\mathbf{e}_i$. These vectors are constant in time to all inertial observers. I will call the inertial reference frame $K$, and this set of vectors $S$.
Secondly, the body frame - a non-inertial reference frame which moves with the rigid body. A second set of orthonormal basis vectors $\mathbf{e}'_j$ move with this frame. I will call this frame of reference $K'$, and this set of vectors $S'$.
The diagram below illustrates this picture (the weird blob is the rigid body!).
Any vector in space can be written as a linear combination of either set of basis vectors: we can write
$$
\mathbf{A} = A_i\mathbf{e}_i = A'_j\mathbf{e}'_j.
$$
In particular, if the vector $\mathbf{A}$ is fixed to the rigid body, then the components $A_i$ are functions of time, but, since the projection of $\mathbf{A}$ onto each of the $\mathbf{e}'_j$ is constant, the $A'_j$ are constant. This holds in both $K$ and $K'$.
Importantly, an observer from either reference frame can use either set of basis vectors. The difference is in how they view the behaviour of each set of vectors: from the perspective of $K$, the basis vectors $\mathbf{e}_i$ are fixed, but the basis vectors $\mathbf{e}'_j$ are functions of time; however, to an observer in $K'$, i.e. from the perspective of the rigid body, the $\mathbf{e}'_j$ are fixed, and it is instead the vectors $\mathbf{e}_i$ which are time dependent.
Thus, when observed from $K'$,
$$
\left(\frac{\text{d}}{\text{d}t}\right)_{K'}\mathbf{e}'_i = \mathbf{0}
$$
so the rate of change of a vector $\mathbf{A} = A'_i\mathbf{e}'_i$ appears to be
$$
\left(\frac{\text{d}\mathbf{A}}{\text{d}t}\right)_{K'} = \dot{A'}_i\mathbf{e}'_i.
$$
In $S$ howver, the $\mathbf{e}'_i$ basis vectors, are functions of time. I show below that, as observed from $K$,
$$
\left(\frac{\text{d}}{\text{d}t}\right)_{K}\mathbf{e}'_i = \boldsymbol{\omega}\times\mathbf{e}'_i\tag{1}
$$
for some time-dependent vector $\boldsymbol{\omega}$, so
\begin{align}
\left(\frac{\text{d}\mathbf{A}}{\text{d}t}\right)_{K} &= \dot{A'}_i\mathbf{e}'_i +A'_i\dot{\mathbf{e}}'_i\\
& = \dot{A'}_i\mathbf{e}'_i + A'_i \boldsymbol{\omega}\times\mathbf{e}'_i\\
& = \dot{A'}_i\mathbf{e}'_i+ \boldsymbol{\omega}\times(A'_i \mathbf{e}'_i)\\
& = \dot{A'}_i\mathbf{e}'_i +\boldsymbol{\omega}\times\mathbf{A},
\end{align}
and thus
$$
\left(\frac{\text{d}\mathbf{A}}{\text{d}t}\right)_{K'} = \left(\frac{\text{d}\mathbf{A}}{\text{d}t}\right)_{K} - \boldsymbol{\omega}\times\mathbf{A}. \tag{2}
$$
This is the crux of the problem: to an observer in the moving frame, every vector appears to have an additional term in its time derivative, when in fact it is the frame itself that is changing.
With this, Euler's equations follow quite naturally. Let $\mathbf{T}$ be the torque on the body, and let $\underline{\mathbf{I}}\boldsymbol{\omega}$ be the angular momentum of the body, where $\underline{\mathbf{I}}$ is the moment of inertia tensor. The key point is that as measured in $K'$, the moment of inertia tensor is constant. (This is not true in $K$, as the distribution of the mass of the body with respect to the $\mathbf{e}_i$ basis changes with time.) Using equation (2), this means that
$$
\left(\frac{\text{d}}{\text{d}t}\right)_{K'}\underline{\mathbf{I}}\boldsymbol{\omega} = \underline{\mathbf{I}}\left(\frac{\text{d}}{\text{d}t}\right)_{K'}\boldsymbol{\omega} = \underline{\mathbf{I}}\left[\left(\frac{\text{d}}{\text{d}t}\right)_{K}\boldsymbol{\omega}-\underbrace{\boldsymbol{\omega}\times\boldsymbol{\omega}}_{\mathbf{0}}\right] = \underline{\mathbf{I}}\left(\frac{\text{d}}{\text{d}t}\right)_{K}\boldsymbol{\omega}.\tag{$\star$}
$$
Finally, the rate of change of angular momentum in the body is equal to the applied torque:
$$
\mathbf{T}= \left(\frac{\text{d}}{\text{d}t}\right)_{K}\underline{\mathbf{I}}\boldsymbol{\omega} = \left(\frac{\text{d}}{\text{d}t}\right)_{K'}\underline{\mathbf{I}}\boldsymbol{\omega}+\boldsymbol{\omega}\times(\underline{\mathbf{I}}\boldsymbol{\omega}) =\underline{\mathbf{I}}\left(\frac{\text{d}}{\text{d}t}\right)_{K}\boldsymbol{\omega} +\boldsymbol{\omega}\times(\underline{\mathbf{I}}\boldsymbol{\omega}).
$$
This is Euler's equation! It holds from the perspective of the inertial frame, but is derived using the basis vectors of the non-inertial frame.
All that remains is to prove equation (1). To do so, it is best to forget about the body frame $K'$, and to think of the $\mathbf{e}'_i$ as simply a set of vectors which move with respect to the $\mathbf{e}_i$, as seen from $K$. The moving basis vectors $\mathbf{e}'_j(t)$ are related to the fixed basis vectors $\mathbf{e}_i$ by a time dependent rigid rotation, which can be described by a matrix $\underline{\mathbf{R}}(t)$. Using Einstein notation, we have
$$
\mathbf{e}'_i(t) = R_{ij}(t) \mathbf{e}_j, \qquad\text{and}\qquad \mathbf{e}_i = R_{ij}^T(t) \mathbf{e}'_j(t). \tag{1}
$$
Since $\underline{\mathbf{R}}$ is a rotation matrix, it is orthogonal, so for all times $t$ it satisfies
$$
\underline{\mathbf{R}}(t)\underline{\mathbf{R}}^T(t) = \underline{\mathbf{I}},
$$
where $\underline{\mathbf{I}}$ is the identity matrix. Differentiating this with respect to time we find
$$
\underline{\dot{\mathbf{R}}}(t)\underline{\mathbf{R}}^T(t)+\underline{\mathbf{R}}(t)\underline{\dot{\mathbf{R}}}^T(t)= \underline{\dot{\mathbf{R}}}(t)\underline{\mathbf{R}}^T(t) + \left(\underline{\dot{\mathbf{R}}}(t)\underline{\mathbf{R}}^T(t)\right)^T = \underline{\mathbf{0}},
$$
so the matrix $\underline{\dot{\mathbf{R}}}(t)\underline{\mathbf{R}}^T(t)$ must be antisymmetric. This implies that there exists some vector $\boldsymbol{\omega} = (\omega_1, \omega_2, \omega_3)$ such that
$$
\underline{\dot{\mathbf{R}}}(t)\underline{\mathbf{R}}^T(t) = [\boldsymbol{\omega}\times]
$$
where the matrix
$$
[\boldsymbol{\omega}\times] = \begin{pmatrix}0 & -\omega_3 & \omega_2 \\ \omega_3 & 0 & -\omega_1 \\ -\omega_2 & \omega_1 &0 \end{pmatrix}
$$
acts on vectors like a cross product as $[\boldsymbol{\omega}\times]\mathbf{A} = \boldsymbol{\omega}\times\mathbf{A}$ (see here if you've not seen this before). For our purposes, it will be most useful to know the components of this matrix in the moving basis $\mathbf{e}'_j$. Using $\boldsymbol{\omega} = \omega'_j\mathbf{e}'_j$, we have
$$
\left(\underline{\dot{\mathbf{R}}}(t)\underline{\mathbf{R}}^T(t)\right)'_{ij} = \mathbf{e}'^T_i\left(\underline{\dot{\mathbf{R}}}(t)\underline{\mathbf{R}}^T(t)\right)\mathbf{e}'_j = \epsilon_{ijk}\omega'_k.
$$
With this, we can calculate the time derivatives of the primed basis vectors:
\begin{align}
\dot{\mathbf{e}}'_i(t) &= \dot{R}_{ij}(t) \mathbf{e}_j \\
&= \dot{R}_{ij} R^T_{jk}\mathbf{e}'_k\\
&= \epsilon_{ikm}\omega'_m \mathbf{e}'_k\\
&= \epsilon_{ikm}(\boldsymbol{\omega}\cdot\mathbf{e}'_m)\mathbf{e}'_k\\
&= \frac{1}{2} \epsilon_{ikm}(\boldsymbol{\omega}\cdot\mathbf{e}'_m)\mathbf{e}'_k - \frac{1}{2}\epsilon_{ikm}(\boldsymbol{\omega}\cdot\mathbf{e}'_k)\mathbf{e}'_m\\
&=\frac{1}{2}\epsilon_{ikm}\boldsymbol{\omega}\times(\mathbf{e}'_k\times\mathbf{e}'_m)\\
&=\frac{1}{2}\epsilon_{ikm}\boldsymbol{\omega}\times(\epsilon_{kml}\mathbf{e}'_l)\\
&=\delta_il\boldsymbol{\omega}\times\mathbf{e}'_l\\
&=\boldsymbol{\omega}\times\mathbf{e}'_i(t).\tag{2}
\end{align}
Thus, as seen from the inertial frame, there exists an instantaneous angular velocity vector $\boldsymbol{\omega}$ around which the primed basis vectors $\mathbf{e}'_i(t)$ rotate.
Finally, here is an alternative derivation of Euler's equation that I am particularly fond of. We proceed by directly by differentiating the moment of inertia tensor. To do this, we align the $\mathbf{e}'_i$ with the principal axes of the body. We denote the principal moment of the body along the $\mathbf{e}'_i$ axis as $I_i$. Since the body is rigid, the pricipal moments are constants, and so we can explicitly write the moment of inertial tensor as
$$
\underline{\mathbf{I}} = \sum_i I_i \mathbf{e}'_i \otimes \mathbf{e}'_i
$$
(summation convention not used here).
The time derivative of $\underline{\mathbf{I}}$ (in $K$) is then
$$
\left(\frac{\text{d}}{\text{d}t}\right)_{K} \underline{\mathbf{I}} = \sum_i I_i\left[\dot{\mathbf{e}}'_i\otimes\mathbf{e}'_i+\mathbf{e}'_i\otimes\dot{\mathbf{e}}'_i\right] = \sum_i I_i\left[(\boldsymbol{\omega}\times\mathbf{e}'_i)\otimes\mathbf{e}'_i+\mathbf{e}'_i\otimes(\boldsymbol{\omega}\times\mathbf{e}'_i)\right].
$$
Now, let $\mathbf{b}$ and $\mathbf{c}$ be arbitrary vectors, and $\underline{\mathbf{A}}$ be a linear operator. Using the definition of the tensor product, we have
$$
\left[(\underline{\mathbf{A}}\mathbf{b})\otimes\mathbf{c}\right]_{ij}=\left(\underline{\mathbf{A}}\mathbf{b}\right)_i c_j = A_{ik}b_k c_j = \left(\underline{\mathbf{A}}\right)_{ik}\left(\mathbf{b}\otimes\mathbf{c}\right)_{kj} = \left[\underline{\mathbf{A}}(\mathbf{b}\otimes\mathbf{c})\right]_{ij}
$$
and
$$
\left[\mathbf{b}\otimes(\underline{\mathbf{A}}\mathbf{c})\right]_{ij}=b_i\left(\underline{\mathbf{A}}\mathbf{c}\right)_j = b_i A_{jk} c_k = \left(\mathbf{b}\otimes\mathbf{c}\right)_{ik}\left(\underline{\mathbf{A}}^T\right)_{kj} = \left[(\mathbf{b}\otimes\mathbf{c})\underline{\mathbf{A}}^T\right]_{ij}.
$$
Therefore,
\begin{align}
\left(\frac{\text{d}}{\text{d}t}\right)_{K} \underline{\mathbf{I}}&= \sum_i I_i\left[(\boldsymbol{\omega}\times\mathbf{e}'_i)\otimes\mathbf{e}'_i+\mathbf{e}'_i\otimes(\boldsymbol{\omega}\times\mathbf{e}'_i)\right] \\
&= \sum_i I_i\left[[\boldsymbol{\omega}\times](\mathbf{e}'_i\otimes\mathbf{e}'_i)+(\mathbf{e}'_i\otimes\mathbf{e}'_i)[\boldsymbol{\omega}\times]^T\right]\\
&=[\boldsymbol{\omega}\times]\underline{\mathbf{I}}-\underline{\mathbf{I}}[\boldsymbol{\omega}\times]
\end{align}
where $[\boldsymbol{\omega}\times]$ is the operator defined in the previous section, and in the last line I have used that this operator is anti-symmetric.
Thus, the time derivative of the angular momentum is
\begin{align}
\mathbf{T} &= \left(\frac{\text{d}}{\text{d}t}\right)_{K} (\underline{\mathbf{I}}\boldsymbol{\omega}) \\
&= \underline{\mathbf{I}}\left(\frac{\text{d}}{\text{d}t}\right)_{K} \boldsymbol{\omega}+\boldsymbol{\omega}\left(\frac{\text{d}}{\text{d}t}\right)_{K} \underline{\mathbf{I}}\\
&=\underline{\mathbf{I}}\left(\frac{\text{d}}{\text{d}t}\right)_{K} \boldsymbol{\omega} + [\boldsymbol{\omega}\times]\underline{\mathbf{I}}\boldsymbol{\omega}-\underline{\mathbf{I}}\underbrace{[\boldsymbol{\omega}\times]\boldsymbol{\omega}}_{\boldsymbol{\omega}\times\boldsymbol{\omega}=\mathbf{0}}\\
&=\underline{\mathbf{I}}\left(\frac{\text{d}}{\text{d}t}\right)_{K} \boldsymbol{\omega} + \boldsymbol{\omega}\times(\underline{\mathbf{I}}\boldsymbol{\omega})
\end{align}
as before.
To explain how orientation can change whilst angular momentum is conserved it is first best to look at a slightly simpler system - a cat in free fall!
Here is a series of photographs taken in $1894$ which shows a cat turning its body to ensure that it lands on its feet.
This gif file illustrates how a cat changes its shape to rotate and yet still to conserve angular momentum.
Finally here is a video of such an event with the cat suffering no harm.
So the key is changing body shape to achieve a rotation whilst conserving angular momentum.
This is shown using a selection of stills from the gymnast video.
First head on.
Arm movements starting in slide $\rm d$ initiate the twisting of the gymnast.
From the side.
Here is a dive executed in the video referenced by the OP.
The diver when on the diving board cannot use it to start a twisting rotation as that rotation could not be removed towards the end of the dive and I think that is also against completion rules.
The Physics of somersaulting and tumbling is explained in an article published in Scientific American.
By moving the arms a diver can start and stop a twist.
The somersault rotation continues from start to finish but before entering the water the diver increases the moment of inertia about a horizontal axis by stretching out thus reducing the speed of rotation.
By timing the entry to perfection and whilst still rotating the diver enter the water with the smallest horizontal profile.
Note the rotation continuing under the water.
This slow motion video of a twisting somersault shows clearly how the arms are used to initiate twisting.
Best Answer
You need to add the spin angular momentum to the orbital angular momentum.
The spin angular momentum is the angular momentum of the body about its centre of mass and the orbital angular momentum is the angular momentum of the centre of mass of the body about the axis defined in the lab frame, shown as a red blob in your diagram.
So for the top left cup, $L_{\rm spin} = I_{\rm \text{CoM cup}}\,\omega_1$ and $L_{\rm orbital} = I_{\text {red blob}}\,\color{red}{\omega_1}$