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For a better understanding of this question, please see the question and answer here.

In $Spin(8)$ there are plenty of copies of $Spin(7)$; consider, for instance, the antiimage of $SO(7)<SO(8)$ by the double cover for any $SO(7)<SO(8)$ obtained by fixing some direction in $\mathbb{R}^8$. So it is true, there are plenty of $Spin(7)<Spin(8)$ (let's call this subgroups primitives).

By the triality automorphism of $Spin(8)$, the $Spin(7)$ subgroups go to other $Spin(7)$ subgroups. But the hot point is that it is not a permutation between primtives subgroups: that is, if we perform triality on a $Spin(7)<Spin(8)$ obtained by fixing $e_1\in\mathbb{R}^8$, then we get a copy $Spin(7)<Spin(8)$ whose proyection onto $SO(8)$ is not a double cover of some subgroup $SO(7)<SO(8)$, but an isomorphism over some $Spin(7)<SO(8)$. And it happens that this $Spin(7)<SO(8)$ may be described as the stabilizer of a 4-form, as described in the link above.

So the corner question is: Why this $Spin(7)$ obtained by performing triality to a primitive $Spin(7)$ should in fact be so strangely described, as the stabilizer of some 4-form? Is there a nice way to understand this fact?

Any idea or suggestion is welcome.

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I think that you want to look at the relevant passages in Spin Geometry by Lawson and Michelsohn, particularly Chapter IV, Sections 9 and 10, where they explain in general how the square of a spinor can be written as a sum of differential forms. This gives a general way to see how having a parallel spinor (in any dimension) defines a set of parallel forms and conversely. This may be the 'nice way to understand this fact' that you seek.

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    $\begingroup$ "and conversely": in general, must every invariant form be the square of some invariant spinor (say, wrt some invariant bilinear/sesquilinear form on complex spinors)? $\endgroup$
    – Paul Reynolds
    Commented Mar 5, 2015 at 11:46
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    $\begingroup$ @PaulReynolds: Sorry, I wrote the above in some haste and it's perhaps not as clear as it could be. I didn't mean to suggest that a parallel form implies a parallel spinor in all dimensions. That's certainly not true in general. What I meant was that, sometimes, the group of linear transformations fixing some set of forms on a vector space can be characterized as the subgroup that fixes the square of a spinor. The reference I mentioned above explains this fully. $\endgroup$
    – Robert Bryant
    Commented Mar 5, 2015 at 12:31
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    $\begingroup$ Right, thanks, of course, e.g. the volume form on any generic holonomy space. I have never really understood what's particularly special about the forms that do come from spinors vs those that don't, but I guess that's another question. $\endgroup$
    – Paul Reynolds
    Commented Mar 5, 2015 at 12:55
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    $\begingroup$ @PaulReynolds the "converse" that you mention has actually been studied recently in arXiv:1911.08658, where the authors characterize which polyforms can be written as the square of a spinor. It turns out that a polyform is the square of a spinor if and only if it satisfies a natural system of algebraic equations in the Kähler-Atiyah bundle of the underlying manifold. $\endgroup$
    – Bilateral
    Commented Mar 21, 2021 at 22:05

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