The result of the product $\vec{a}\times(\vec{b}\times\vec{c})$ should be orthogonal to $\vec{a}$, and the right-hand is indeed orthogonal to $\vec{a}$. If you compute it's dot product with $\vec{a}$ $(\vec{a}\cdot\vec{b})(\vec{a}\cdot\vec{c})-(\vec{a}\cdot\vec{c})(\vec{a}\cdot\vec{b})$ there is a symmetry that causes cancellation to $\vec{0}$. So this provides one way to remember the identity. $\vec{a}$, having the "least symmetric" role in $\vec{a}\times(\vec{b}\times\vec{c})$ is involved with both dot products on the right.
The identity should involve that minus sign, since swapping the roles of $\vec{b}$ and $\vec{c}$ should negate the resultant cross product vector. Also, as already noted, the presence of that negative sign helps confirm that $\vec{a}$ is orthogonal to the cross product.
At a more basic level, the result of a cross-product is a vector. So the right-hand side can be a linear combination of vectors (and not, say, of dot products). Here, we combine two vectors using scalar weights that are each obtained from dot products.