Is this statement correct or incorrect: Moving objects undergo length contraction because they appear in space-time cross-section Why or why not?
This is closely related to another question I posited here, Does it make sense to say that objects moving at relativistic velocities appear in space-time cross-section?
What I mean is the following: Saying that moving objects appear in space-time cross-section along a space-like hypersurface (the observed 3D space view at a given moment) in the geometry of SR implies that they must undergo length contraction. 
Motivation: I used a similar statement in a couple of videos I made on length contraction and the length contraction paradox and got mixed responses. My intent was in part an unorthodox approach that highlights the strong connection between length contraction and relativity of simultaneity beyond the common examples. Although the videos use strictly standard special relativity, somebody commented to me that such statements sound unprofessional and likely leave the impression of crackpottery. I posted a similar question on Reddit/r/Physics and got mostly the same reaction, which to me is a big surprise. 
Are there really any conceptual objections to this that I overlooked? Or is it just a matter of the audience being unprepared for it?
Note: The original version posted on Reddit was "Moving objects appear to undergo length contraction because they are perceived in space-time cross-section". A major objection over there was that using words like "appear" and "perceive" suggest I consider length contraction some sort of optical illusion dependent on the observer. To the contrary. I never intended to debate that it is a real, measurable effect. Only meant to emphasize that length contraction appears differently in different frames, as demonstrated by measured coordinates. Rephrased here to eliminate this sort of misunderstanding.
Let me refine and clarify please:
Let us limit the question to special relativity only, for convenience and clarity. Let us also leave aside optical effects in SR, including doppler, beaming, Terrell-Penrose, etc. In other words, the question does not refer to the optical observation of length contraction by a human observer or a camera. Let us discuss only in terms of measured coordinates.
Let object A move at relativistic velocity relative to an inertial frame O. In 4D space-time (Minkowski diagram) the space view of O at any given moment of its own time is a space-like hyperplane. What O observes of A at any time is the cross-section of A's world-tube by O's corresponding constant-time hyperplane. In this cross-section A undergoes length contraction. Geometrically speaking, the length contraction is a consequence of the fact that O's constant-time hyperplanes "slice" A's world-tube at a different angle than do A's own hyperplanes of constant proper-time (corresponding to the 3D space view in its rest frame).
In view of this 4D representation, is it ok to say that, as seen in O, A undergoes length contraction because it appears in a different space-time cross-section than it does in its rest frame?
 A: It's the same question but now we've specialized on length contraction. Right on.
Why we can say what we're saying
You can work this out from Lorentz transforms if you want, but Lorentz invariants let you simply write it down: a particle has an invariant 4-velocity $\gamma~[c, \vec v]$ and we can form the inner product of this with the local 4-positions $[ct, \vec r]$ to find the equation for that other particle's spacelike hyperplane in my coordinates, as in their coordinates this 4-velocity is $[c, 0]$ and picks out only the time component $c^2 ~ t'$ in the local coordinates. Their "present moments" therefore have the form
$$c t - v_x ~ x - v_y ~ y - v_z ~ z = C = c^2 ~ t' / \gamma.$$
Therefore, if you have a spacelike hyperplane $a_\mu x^\mu = A$, the parameters $u^i = -c~a_i / a_0$ tell you the 3D components of $\vec v$ directly. Since you have $\vec v$ you have everything you need to derive the Lorentz transform that makes that hyperplane constant-time. However, if the hyperplane is too tilted, you will presumably get a $\vec v$ with magnitude greater than $c$, which among other things will be unphysical.
Interpretation nuances
While we have everything we need to determine the Lorentz boost, we have to be really careful with how we interpret the geometry here.
One example: the light cone $c t - x - y - z = 0$ must look like a light cone in all other reference frames due to the mathematics of Lorentz transforms. But if you remember your conic sections, the relevant conic section is an ellipse, a circle that's been stretched along one axis. What we call light cones are actually uniformly expanding bubbles at speed $c$, but the point still holds: the shape that is seen in the hyperplane is not a sphere, but is linearly extended in some direction. This direction, indeed, is $\vec v$, and the extension of the sphere in our version of the hyperplane is corrected for exactly by the Lorentz contraction of the hyperplane when you make the Lorentz transform into the local coordinates. 
In other words, it is too simplistic to say "A undergoes length contraction because it appears in a different space-time cross-section than it does in its rest frame." In fact we see that any "vertical tube" (i.e. a sphere of constant radius staying in one place over time) has a lengthened cross-section in the hyperplane, which must be corrected for by the relativistic $\gamma$ to make it length-contract. Instead we have to play a game where this timelike hyperplane intersect a light cone which intersects all of the "perpendicular" (i.e. to $\vec v$) spatial dimensions at the same time, but intersects the other dimension $x$ at different points in time. The light cone's conic section is bigger, and when we scale it down to a circle, then the tube has contracted in the $x$ direction.
