I would like to take a pair $(\zeta^2,v)$ and find a pair $(\zeta^3,w)$ where there is an embedding $e: \zeta^2 \hookrightarrow \zeta^3$ such that $v=e^*w.$
Define the manifold $\zeta^2:=(0,1)^2$ and vector field $v:=\langle x\log x, -y\log y\rangle,$ for $x,y \in(0,1).$
Due to the comment below...$w$ should be a 3-vector field accumulating to each of the eight vertices and non-vanishing elsewhere. (source is at $(1,0,0)$ and sink is at $(0,1,1)$).
Attempt:
Lift $(\zeta^2,v)$ onto a parallelogram so that it's embedding in $\zeta^3$ spans the long diagonal of $\zeta^3,$ as opposed to sitting at the base of the cube (approximate sketch above).
Then $w$ should restrict onto the lifted $(\zeta^2,v).$ I'm thinking of this parallelogram as being bent/curved whilst the $(0,1,1)$ and $(1,0,0)$ points stay fixed. On each of these templates in the parameter space will be $v$ but it will sit on manifolds of varying curvatures.
Finally take the collection of these vector fields $v_i$, to get $w.$
This is only an attempt and may not be the correct way to approach the problem.
How do we arrive at the pair $(\zeta^3,w)?$
If for example $\zeta^2$ is defined to be the closed 2-disk, $D^2,$ then one example for $v$ is $v(p) := \frac{1}{r-\|p\|}\frac{\partial}{\partial x}(p).$ Where $r$ is the radius of the disk.
$e(x,y)=(x,y,0)$ (possibly $ (x,y,1))$ and define $$ W(x,y,z)=\bigg(x{\rm log}x,-y{\rm log}y,-z{\rm log}z \bigg) $$
Then $de\ \frac{\partial }{\partial x} = (1,0,0),\ de\ \frac{\partial }{\partial y} = (0,1,0) $ so that $$ (de\ V)(x,y,0)=W(x,y,0) $$ where $V=(x{\rm log}x,-y{\rm log}y)$.