Helmholtz decomposition vs Laplacian vector fields on $\mathbb{T}^3$?

The correct generalization to $\mathbb{T}^3 = (\mathbb{R}/\mathbb{Z})^3$ is the Hodge decomposition. Every vector field on $\mathbb{T}^3$ is uniquely expressible as $u = \nabla \times v + \nabla w + h$, where $\nabla \cdot h = \nabla \times h = 0$ is a Laplace vector field in your terminology. It is the summands that are unique in the decomposition, obviously $v \mapsto v + \nabla \chi$ and $w \mapsto w + C$ with constant $C$ produce the same $u$. If $\nabla \cdot u = 0$ then the $\nabla w$ term can be dropped, same for $\nabla \times v$ when $\nabla \times u = 0$. There are topological obstructions to expressing the $h$ term as either a curl or a divergence, but $h$ belongs only to a finite dimensional space. In fact, the space of Laplace vector fields is isomorphic to the de Rham cohomology $H^1(\mathbb{T}^3; \mathbb{R}) \cong H^2(\mathbb{T}^3; \mathbb{R}) \cong \mathbb{R}^3$, which in this case consists explicitly of the span $\langle \partial_x, \partial_y, \partial_z \rangle$, using the standard $(x,y,z)$ coordinates on $\mathbb{R}^3$ (before quotienting by $\mathbb{Z}^3$).

Above, I have simply translated the general statement of the Hodge decomposition to $\mathbb{T}^3$ and the language of the OP. The main observation is that, using appropriate Hodge dualization, the sequence of operators (gradient, curl, divergence) becomes the sequence of de Rham differentials $d$ acting on differential forms (dualizing once more, the same sequence becomes the sequence of de Rham codifferentials $\delta$ on forms). The harmonic forms (of degree 1 or 2) of Hodge theory then become the Laplace vector fields of the OP (harmonic forms of degree 0 or 3 become constant functions).