Aren't you simply asking for "Cartesian product"? The Cartesian product of $m$ convex polytopes each with $v$ vertices and described by $h$ hyperplane inequalities in $\mathbb R^t$ is a polytope with $v^m$ vertices described by $mh$ inequalities and lives in $\mathbb R^{tm}$.

Very nice answer. Just wanted to add that the "rectified 5-simplex" is sometimes called the "2nd hypersimplex of dimension 5". Hypersimplices of dim $d$ are the slices of the $(d+1)$-cube obtained by intersecting with the hyperplane $\sum x_i=k$. In our case $k=2$, hence the name "second". ($k= 1$ gives the standard simplex). Note: Some days ago I inadvertently placed this as a comment to my answer. It probably looked weird, to say the least, that I called my own answer very nice...

According to grdb.co.uk/search/toricf3c there are 416 reflexive $3$-polytopes of volume 20 and with 12 boundary points. You need to find one whose boundary can be triangulated as an icosahedron with the triangulation being regular (aka coherent, aka projective). Among the 416 examples I wouldd expect there to be at least one... As an example, the fan in Panov's answer below comes from triangulating the boundary of a cube octahedron, which is the reflexive polytope with ID 12645 in grdb.co.uk/search/toricf3c

(continued) In general, you can still say that every convex combination of extremal sections (regular triangs) is a section, and that they cover all points in the secondary polytope. But you loose uniqueness: for each point $x$ in the secondary polytope you get as many sections as ways there are of writing $x$ as a convex combination of vertices. Ok, one way to make the section unique (but not very canonical) is to choose a triangulation of the secondary polytope, and for each $x$ use the convex combination coming from the vertices of the simplex containing $x$...

I do not see a natural way of choosing one section. Let $A$ be the vertices of a quadrilateral. $\pi$ projects a tetrahedron $T$ to $\conv(A)$ and there are two extremal sections, the "upper" and "lower hull" of $T$ (wrt $\pi$), corresponding to the two vertices of the secondary polytope, a segment. Certainly, you can say that any convex combination of these two extremal sections is a section, and they give a canonical representative for each point in the secondary polytope. But you are able to do this only because your secondary polytope was a simplex (continues below).