Wikipedia defines the geometric notion of a lattice as a discrete subgroup of $\mathbb{R}^n$ (i.e. a subgroup isomorphic to $\mathbb{Z}^n$. This can be viewed as the span of a basis for $\mathbb{R}^n$ over integer coefficients.
If we have such a lattice $L$, with elements $x,y$, we could define an ordering by $x \leq y$ if and only if $y = x + span_\mathbb{N}(\beta) $, where $\beta$ is the basis which we can identify with $L$. This clearly defines a preorder on $L$, and at least in the two dimensional case, geometrically it seems like should be a lattice.
Does this construction always give us an (order-theoretic) lattice? If not, will some other construction?
Sure. As you said, a lattice $L$ is a subgroup of $\mathbb{R}^n$ isomorphic to $\mathbb{Z}^n$, and $\mathbb{Z}^n$ admits a lattice structure, where meet and join are coordinate-wise min and max. So you can put the same structure on $L$ via the bijection with $\mathbb{Z}^n$.
This agrees with your definition: If $x = (x_1,\dots,x_n)$ and $y = (y_1,\dots,y_n)$ in $\mathbb{Z}^n$, then $x\leq y$ in the lattice order on $\mathbb{Z}^n$ if and only if $x_i \leq y_i$ for all $i$, if and only if $y = x + z$, where $z$ is some element of $\mathbb{Z}^n$ with non-negative coordinates, i.e. some linear combination of the standard basis $e_i$ for $\mathbb{Z}^n$ with coefficients in $\mathbb{N}$. And the standard basis for $\mathbb{Z}^n$ corresponds to your basis $\beta$ for $L$ under the chosen isomorphism.