Do you mean
\vec{V}\nabla^2\vec{V}? That is commonly called "del squared". "gradient" is specifically \nabla f where f is a scalar valued function. "curl f" is \nabla\times \vec{f} where f is a vector valued function, and "div f" is \nabla\cdot \vec{f}.
Now, \nabla^2= \frac{\partial^2}{\partial x^2}+ \frac{\partial^2}{\partial y^2}+ \frac{\partial }{\partial z^2} (in Cartesian coordinates) is also, strictly speaking, applied to a scalar function but it is not uncommon to see it applied to a vector meaning \nabla^2 applied to each component.
That is, if
\vec{v(x,y,z)}= f(x,y,z)\vec{i}+ g(x,y,z)\vec{j}+ h(x,y,z)\vec{k} then
\nabla^2 \vec{v}(x,y,z)= \nabla^2f \vec{i}+ \nabla^2g\vec{j}+ \nabla^2h \vec{k}
a vector. But to say what "\vec{v}\nabla^2\vec{v}" means we would still have to know what kind of vector product that is- dot product or cross product?
If that is dot product and \vec{v}(x,y,z)= f(x,y,z)\vec{i}+ g(x,y,z)\vec{j}+ h(x,y,z)\vec{k} then
\vec{v}\cdot \nabla^2\vec{v}= f(x,y,z)\nabla^2 f(x,y,z)+ g(x,y,z)\nabla^2 g(x,y,z)+ h(x,y,z)\nabla^2h(x, y, z)
If, instead, it is the cross product then
\vec{v}\times\nabla^2\vec{v}= \left(g(x,y,z)\nabla^2h(x,y,z)- h(x,y,z)\nabla^2g(x,y,z)\right)\vec{i}- \left(f(x,y,z)\nabla^2h(x,y,z)- h(x,y,z)\nabla^2f(x,y,z)\right)\vec{i}+ \left(f(x,y,z)\nabla^2g(x,y,z)- g(x,y,z)\nabla^2f(x,y,z)\right)\vec{k}
