All you need is the exterior derivative and the Hodge star; you can already anticipate this because the expression you're looking at is $\nabla \times (\mathbf{w}\times \mathbf{u})$, and the curl is really a manifestation of $d$, the metric tensor and the Hodge star, while the cross product is "essentially" the wedge product of two 1-forms once you make the relevant identifications I describe below. You shouldn't expect any Lie derivatives $L_\mathbf{u}$ because they measure the rate of change of tensor fields as you flow them along $\mathbf{u}$, so in the classical vector calculus this would somehow manifest as a dot product with $\mathbf{u}$.
Just so we're on the same page, we equip $\Bbb{R}^3$ with its usual metric tensor $g$, the standard orientation, and let $\star$ be the associated Hodge star. Also, let $g^{\flat}$ and $g^{\sharp}$ be the associated musical isomorphisms. Given any vector fields $\mathbf{w},\mathbf{u}, \mathbf{v}$, we have the following definitions:
The cross product is $\mathbf{w}\times \mathbf{u}:= g^{\sharp}[\star (g^{\flat}(\mathbf{w})\wedge g^{\flat}(\mathbf{u}))]$. i.e given our vector fields, we convert them into $1$-forms, and take their wedge product. This yields a $2$-form, so we use $\star$ to convert this into a $1$-form, then finally use $g^{\sharp}$ to convert this into a vector field again.
The curl is $\nabla \times \mathbf{v}:= g^{\sharp}[\star d g^{\flat}(\mathbf{v})]$, i.e we convert the vector field into a $1$-form, take the exterior derivative, which gives a $2$-form, then use $\star$ to make this a $1$-form, then use the metric to get a vector field again.
If you denote $\alpha := g^{\flat}(\mathbf{u})$, and $\beta=d\alpha$, then you should find that $g^{\flat}(\mathbf{w})=g^{\flat}(\nabla \times \mathbf{u}) = \star d\alpha = \star \beta$, by directly plugging into the above definitions (and recalling that $g^{\flat}$ and $g^{\sharp}$ are inverses).
So, if we start with the expression $\partial_t {\bf{w}} + \nabla \times( {\bf{w}} \times {\bf{u}})$, and apply $g^{\flat}$ to make this a $1$-form, then you should find by unwinding the definitions that
\begin{align}
g^{\flat}\left(\partial_t {\bf{w}} + \nabla \times( {\bf{w}} \times {\bf{u}})\right) &= \partial_t(\star d\alpha) + \star d\left((\star d\alpha)\wedge \alpha\right)
\end{align}
If we now apply $\star$ to both sides of this equation, and note that $\star\star = (-1)^{k(n-k)}\text{identity}$, where $k$ is the degree of the form and $n$ is the dimension of the space, it follows that for $n=3$ and $k\in\{0,1,2,3\}$, this is always the identity. Hence,
\begin{align}
\star g^{\flat}\left(\partial_t {\bf{w}} + \nabla \times( {\bf{w}} \times {\bf{u}})\right) &= \partial_t( d\alpha) + d\left((\star d\alpha)\wedge \alpha\right)\\
&\equiv \partial_t\beta + d\left((\star \beta)\wedge \alpha\right).
\end{align}
So, this is how you can convert your vector equation into an equation regarding differential forms:
\begin{align}
\partial_t\beta + d\left((\star \beta)\wedge \alpha\right)&= 0, \quad
\text{where $\alpha := g^{\flat}(\mathbf{u})$, $\beta=d\alpha$}.
\end{align}
