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Plucker relations are a statement that if you take arbitrary contraction of $w$ by $k-1$ elements of the dual space and then wedge it with $w$, you will get $0$. If you have a tensor $w$ that satisfies this, then consider the space of all such contractions. Take a basis of this space $A$ and complete it to the basis of $V$. Write $w$ as a linear combination of wedges of subsets of this basis. If there is a nonzero coefficient at something other than ones in $A$, then one can get a contraction that is not in $A$. Thus $w$ is up to a scalar the top wedge of $A$. Is this what you need?
I think that it can be subtle. For example, for conics in $A^2$, parabola intersects every line in at least one point, while the nondegenerate ones have lines that are tangent to it at infinity and thus have zero intersection points in $A^2$.
The concept of linear subspace is not invariant under automorphisms of affine space, so the question is ill-formed. It appears however, that you are thinking about a specific embedding of $X$ into $A^n$, then you can take the closure of $X$ under standard $A^n\subset P^n$ -- its degree is the upper bound you seek. Lower bounds appear more subtle, even in algebraically closed field case.
No, not at all: the algorithm does not use the function. Rather, it is based on the rotation idea. First, you pick ANY decomposition into $x_ix_i^T$. Then at each step, you take the "smallest" and the "largest" indices as far as $x_i^TAx_i$ are concerned and consider the rotations above. For some choice of $\cos \alpha$ one will get the correct value of $x^TAx$ (since at $\alpha=0$ the value is too small at $\alpha=\frac \pi 2$ it is too big; also, it is not a terribly hard equation). You replace the two $x$-s by two $y$-s thus increasing the number of indices for which the $x_i^TAx_i$ is OK.
This can be made into an effective algorithm by picking at each step indices with minimum and maximum $x_i^TAx_i$ and performing a rotation to make one of the values the desired $\frac 1R {\rm Trace}(AX)$.
Have you tried it for $2x2$ matrices? It may simply be a mistake in the paper. Wouldn't the rank one decomposition of a general positive definite symmetric $2x2$ matrix be essentially unique (the axis of the ellipse)? Then there is no way to get the desired property for arbitrary $A$.
Once $1^n$ appears, it is going to propagate itself roughly as much as the length of the sequence. Specifically, after $t$ steps there will be roughly $(3/2)^t$ instances of $1^n$. So heuristically, the probability of $1^n$ converted to $1^{n+1}$ is give or take the same. This strongly suggest that $1^{n+1}$ would eventually appear. Unfortunately, this eventually is of the magnitude $1/N$ where $N$ is the length of the sequence when $1^n$ appeared. So $1^7$ is likely still far far off. There are other issues, as r.e.s. pointed out, but they appear to be minor.
Regarding the 2x2x2 case: It might be a mess to do by hand, but there is a way of coding this. Specifically, have the computer go through all possible angle permutations of 8 triangles (there are $6^7$ possibilities, since you can fix the angles of the first one). You can use symmetry of the problem to reduce this a bit further, if needed. This gives you equations among the squares of the sides $|X_iY_j|^2,|Y_iZ_j|^2,|Z_jX_i|^2$. Add to these the equations of planarity (mathworld.wolfram.com/Cayley-MengerDeterminant.html) and see if the only solution is all zeros by Groebner bases.
I have a feeling that the number of facets of the total order polytope grows faster than a polynomial in $n$. As a result, the polytope can not be described by relations with a bounded set of indices.
