Your $\vee$ is essentially multiplication of polynomials. The variety of tensors $x_1 \vee \dotsb \vee x_m$ corresponds to polynomials that factor as products of linear factors. Points of the (projective) variety correspond to hyperplane arrangements. Dually, they correspond to *cycles* of $m$ points (in the dual projective space).

This variety has various names including the **Chow variety**, the **variety of completely decomposable forms**, and the **split variety**. For example Hirotachi Abo calls it the "variety of completely decomposable $d$-forms" (his $d$ is your $m$) and denotes it $\operatorname{Split_d}(\mathbb{P}^n)$ (also, his $n$ is your $n+1$). Douglas A. Torrance calls it the Chow variety and uses the same notation, $\operatorname{Split_d}$. Yongui Guan calls it the Chow variety and denotes it $\operatorname{Ch}_d$. There are plenty, plenty more references which you can find by searching for those terms.

And it is indeed a variety. If you buy the correspondence with zero-dimensional cycles, that's pretty clearly closed–the points might move around and collide, but it's still a zero-cycle. If you want the equations, Guan seems to have some information; that's just what I found online. Better references would be Gelfand-Kapranov-Zelevinsky and Landsberg's book if you have access to them.

Rank with respect to this variety is called Chow rank, split rank, decomposable rank, etc. Catalisano, et al say that for a monomial $M$, an $M$-decomposition of a form $F$ (where $\deg(F)=\deg(M)$) is an expression for $F$ as a sum of terms given by evaluating $M$ at linear forms. Waring rank is $x_1^d$-rank, and Chow (or split, or decomposable) rank is $(x_1 \dotsm x_d)$-rank.

If I may also respond to one of the comments: Indeed, Comon's conjecture was that the rank and symmetric rank of symmetric tensors would be equal. Symmetric rank is the same as Waring rank. Shitov gave a counterexample to that conjecture. The analogous conjecture for border ranks, that the border rank and border symmetric rank (border Waring rank) of a symmetric tensor are equal, seems to be still open (as far as I know, as of September 2020).

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