This is a reply to a PM, but I'm posting it here 'cos it seems to be generally relevant:
For the vertex figure of the tesseract, it is easiest to look at the vertex-first projection (the one with the rhombic dodecahedron envelope) instead of the usual cell-first projection (the cube-within-a-cube):
The vertex in yellow, at the center of the rhombic dodecahedron, is the vertex closest to the 4D viewpoint. You can see that it has 4 edges joined to it in tetrahedral symmetry. When you cut it with a hyperplane, you can imagine that the hyperplane intersects with the edges, say, half way. Then you just connect the intersection points with each other, and you get a tetrahedron.
In fact, if you carefully trace the edges as you go, you can actually derive all the shapes that the tesseract forms when it intersects with a hyperplane vertex-first. I don't have the diagrams scanned into my computer, but I derived all these figures by hand just by tracing them on the rhombic dodecahedron projection, before I wrote a program to do it for me:
The way to derive it is easy: start at the yellow vertex, then send out 4 points along the 4 edges, moving at the same speed. From the start until they reach the 4 vertices at the corners, they form a tetrahedron. Now at the corners, each point splits into 3 points moving along edges that follow those 4 vertices. That makes it 12 points, and if you trace them out, they form a truncated tetrahedron. Keep going and eventually these points will merge in pairs at 6 vertices, forming an octahedron. Now they split up again in twos, but this time along the two other edges that haven't yet been traversed, thus forming a truncated tetrahedron in dual orientation. Keep going and eventually the points merge in 3's at the top and form a tetrahedron in dual orientation.
Finally, there are 4 edges I omitted from this image, because they lie on the far side of the tesseract; but they essentially connect those last 4 points to the center of the projection, but this time not to the nearest vertex but to the farthest vertex (which coincides with the nearest in projection). Tracing the points along these last 4 edges give you a shrinking tetrahedron that eventually becomes a point again.