A good question. Since the cross product ##\vec v \times \vec w## is a vector orthogonal to both ##\vec v## and ##\vec w##, the product ##\vec v \times (\vec w \times \vec u)## is a vector that is orthogonal to both ##\vec v## and ##\vec w \times \vec u##. As such, it lies in the plane spanned by ##\vec w## and ##\vec u## (since ##\vec w## and ##\vec u## are orthogonal to ##\vec w \times \vec u##). Now, this becomes particularly interesting when ##\vec u = \vec v##, since the result must be then be a linear combination of ##\vec v## and ##\vec w## at the same time as it is orthogonal to ##\vec v##, which leaves only one possible direction. Let us see what use we can put this to.
Consider the case when ##\vec v = \vec n## is a normalised vector of length one, i.e., ##\vec n^2 = 1## and take the triple product
$$
\vec n \times (\vec w \times \vec n) = \vec w (\vec n^2) - (\vec n \cdot \vec w) \vec n = \vec w - (\vec n \cdot \vec w) \vec n,
$$
where we have applied the bac-cab rule. Rearranging this leads to the relation
$$
\vec w = \vec n (\vec w \cdot \vec n) + \vec n \times (\vec w \times \vec n).
$$
You may recognise the first term on the right as the projection of ##\vec w## on ##\vec n##. As both terms add up to ##\vec w##, it should therefore be clear that ##\vec n \times (\vec w \times \vec n)## is the component of ##\vec w## orthogonal to ##\vec n##. One common use of the vector triple product is therefore to find the orthogonal component of a vector with respect to another.