Is there an easy way to show that
$$\vec u\times(\nabla \times \vec v)+\vec v\times (\nabla\times \vec u)+ (\vec v\cdot \nabla)\vec u+(\vec u \cdot \nabla )\vec v$$
is equal to
$$(\vec u \times \nabla)\times\vec v+(\vec v\times \nabla)\times \vec u+\vec u (\nabla \cdot v)+\vec v(\nabla \cdot u)$$
other than straight forward computation using components? I know that both are ways to find $\nabla(\vec u \cdot \vec v)$ but I am guessing there must be a way to use identities so that all the computational work is avoided.
Applying the ideas below to $u \times (\nabla \times v)$ and $(u \times \nabla) \times v$ (as is done on my post here for the second term) yields the identities $$ u \times (\nabla \times v) = u \cdot (\nabla v) - (u \cdot \nabla) v,\\ (u \times \nabla) \times v = u \cdot (\nabla v) - u (\nabla \cdot v). $$ With that, we can simplify your first expression as $$ u\times(\nabla \times v)+v\times (\nabla\times u)+ (v\cdot \nabla) u+( u \cdot \nabla ) v =\\ [u\times(\nabla \times v) + (u \cdot \nabla )v]+ [v\times (\nabla\times u) + (v \cdot \nabla)u]= \\u \cdot (\nabla v) + v \cdot (\nabla u). $$ We can simplify the second expression similarly.
Partial answer:
Write $v = (v_1,v_2,v_3)$, $f = (f_1,f_2,f_3)$, and take $e_1,e_2,e_3$ to be the canonical basis vectors (that is, $i,j,k$). We can write $f = \sum_{j=1}^3 f_j e_j$, and $$ \nabla \cdot f = \sum_{i=1}^3 e_i \cdot \frac{\partial f}{\partial x_i}, \qquad \nabla \times f = \sum_{i=1}^3 e_i \times \frac{\partial f}{\partial x_i}, \\ (v \cdot \nabla)f_j = \sum_{i=1}^3 (v \cdot e_i) \frac{\partial f_j}{\partial x_i}, \qquad (v \times \nabla)f_j = \sum_{i=1}^3 (v \times e_i) \frac{\partial f_j}{\partial x_i},\\ (v \times \nabla) \times f = \sum_{i=1}^3 \sum_{j=1}^3 [(v \times e_i) \times e_j] \frac{\partial f_j}{\partial x_i} $$ With that said, we can write the first expression as $$ \sum_{i=1}^3\sum_{j=1}^3 \left[u \times (e_i \times e_j)\frac{\partial v_j}{\partial x_i} + v \times (e_i \times e_j)\frac{\partial u_j}{\partial x_i} + \left(v_i \frac{\partial u_j}{\partial x_i} + u_i \frac{\partial v_j}{\partial x_i}\right)e_j \right], $$ and the second expression as $$ \sum_{i=1}^3 \sum_{j=1}^3 \left[(u \times e_i) \times e_j\frac{\partial v_j}{\partial x_i} + (v \times e_i) \times e_j\frac{\partial u_j}{\partial x_i} + \left( u_i \frac{\partial v_j}{\partial x_j} + v_i \frac{\partial u_j}{\partial x_j}\right) e_i \right]. $$ The rest can be resolved using vector algebra identities.