The cross-product correction factor shows up any time we take the time derivative of a vector described in a moving frame.
When I teach this topic in my classes, I generally introduce it when talking about taking time derivatives of position vectors, because it's easier to visualize in that situation:
Start by considering the positions of two points $A$ and $B$, denoted by vectors $\vec{r}_{A}$ and $\vec{r}_{B}$. We can then take the relative position of $B$ with respect to $A$ as a vector $\vec{r}_{B/A} = \vec{r}_{B}-\vec{r}_{A}$ such that $\vec{r}_{B} = \vec{r}_{A} + \vec{r}_{B/A}$.
Now let's say that the vector $\vec{r}_{B/A}$ is fundamentally defined in a frame whose orientation is encoded by a rotation matrix $R_{AB}$, such that $\vec{r}_{B/A}=R_{AB}\ \vec{r}_{B/A}'$ (where $'$ indicates a vector defined in the rotated frame).
If we take the time derivative of the position of $B$, we then get
\begin{equation}
\frac{d}{dt} \vec{r}_{B} = \frac{d}{dt} \vec{r}_{A} + \left(\frac{d}{dt} R_{AB}\right)\ \vec{r}_{B/A}' + R_{AB} \left(\frac{d}{dt} \vec{r}_{B/A}'\right).
\end{equation}
The time derivative of a rotation matrix is the product of its rotational velocity with itself, $\frac{d}{dt} R_{AB} = \vec{\Omega}_{AB} \times R_{AB}$, such that the total time derivative can be regrouped as
$$
\frac{d}{dt} \vec{r}_{B} = \frac{d}{dt} \vec{r}_{A} + \vec{\Omega}_{AB} \times \left(\vphantom{\frac{d}{dt}}R_{AB}\ \vec{r}_{B/A}'\right) + R_{AB} \left(\frac{d}{dt} \vec{r}_{B/A}'\right).
$$
If you think of $A$ and $B$ as being points on a piece of paper, the first term in the expansion says how the paper is moving at point $A$, which transfers to how it is moving at point $B$. The second term describes how rotation of the paper makes point $B$ orbit around $A$, and the third term describes how point $B$ moves on the paper relative to point $A$.
For your angular momentum problem, you can think of the angular momentum as a "free vector", i.e., saying that we only want the $\vec{r}_{B/A}$ component and not the $\vec{r}_{A}$ component. When we take the time derivative, we then get the latter two terms from the full equation, and thus the "time derivative as seen in the frame" plus the "frame-rotation cross product" terms.
In the standard expression from within the body frame, we then multiply all the terms by ${R}^{-1}_{AB}$. By a bit of algebra that I won't go into right now, it works out that
$$
{R}^{-1}_{AB}\left[\vec{\Omega}_{AB} \times \left(\vphantom{\frac{d}{dt}}R_{AB}\ \vec{r}_{B/A}'\right)\right] = \vec{\Omega}_{AB}' \times \vec{r}_{B/A}'
$$
so that
$$
{R}^{-1}_{AB} \left(\frac{d}{dt} \vec{r}_{B}\right) = {R}^{-1}_{AB}\left(\frac{d}{dt} \vec{r}_{A}\right) + \vec{\Omega}_{AB}' \times \vec{r}_{B/A}' +\frac{d}{dt} \vec{r}_{B/A}',
$$
in which the last two terms are the standard expression for the derivative of a free vector expressed in a rotating set of coordinates.
