Dgdrxrt

In the tetrahedral-octahedral honeycomb, each vertex seems to be incident to 6 octahedra and 8 tetrahedra:

Such simple combinatorial fact is probably well-known, or perhaps even obvious. However, coming from a different field, I struggle to justify it mathematically. I thought perhaps one can read this from the Schläfli symbols notation or the Coxeter diagram, but did not succeed at that.

How would one go at arriving at this result if one does not want to use the rigorous method of counting colorful solids in a Wikipedia picture?

You can check that solid angles of those polyhedra add up to $4pi$. We have (see here for details):

$$
Omega_tetr=2pi-6arcsinsqrt2over3,
quad
Omega_oct=2pi-8arcsinsqrt1over3.
$$
Hence, when 6 octahedra and 8 tetrahedra meet at a vertex, they cover a solid angle given by:
$$
Omega_tot=8Omega_tetr+6Omega_oct=
28pi-48left(arcsinsqrt2over3+arcsinsqrt1over3right).
$$
But the angles between parentheses are complementary (the squares of their sines add up to $1$), hence:
$$
Omega_tot=28pi-48piover2=4pi.
$$

From the Wikipedia page referred to in the question:

For an alternated cubic honeycomb, with edges parallel to the axes and with an edge length of $1$, the Cartesian coordinates of the vertices are: (For all integral values: $i,j,k$ with $i+j+k$ even)

$(i, j, k)$

In this construction of a tetrahedral-octahedral honeycomb, vertex $(i, j, k)$ is incident on $12$ edges, given by the $12$ vectors $(0, pm1,pm1)$, $(pm1, 0,pm1)$, $(pm1, pm1, 0)$, lying in the $3$ rectangular Cartesian coordinate planes meeting at $(i, j, k)$:

The $12$ vertices adjacent to $(i, j, k)$ are marked here in blue. The $6$ rectangular Cartesian coordinate semi-axes passing through $(i, j, k)$ are terminated by red dots, marking the centres of the $6$ octahedra incident on $(i, j, k)$.

Vertex $(i, j, k)$ is incident on $8$ cubes, of side length $1$, belonging to the underlying cubic honeycomb. In each of these cubes, $3$ edges of the tetrahedral-octahedral honeycomb extend diagonally across the $3$ faces of the cube meeting at $(i, j, k)$. The far ends of these $3$ diagonals, together with $(i, j, k)$ itself, constitute the vertices of one of the $8$ tetrahedra incident on $(i, j, k)$.

In the next diagram, I have shaded in the triangular faces of these tetrahedra opposite to vertex $(i, j, k)$:

In the next picture (taken from a different point of view), I have instead shaded in the square cross-sectional slices of the $6$ octahedra incident on $(i, j, k)$ (centred on the $6$ red dots in the first picture):

As can be seen from the last image, the $12$ vertices of the tetrahedral-octahedral honeycomb adjacent to vertex $(i, j, k)$ are the vertices of a cuboctahedron. For more information about this Archimedean solid, and for images of higher quality, see for example The cuboctahedron | Hexnet, or Cuboctahedron - Wikipedia.

From the Wikipedia page just referred to:

The Cartesian coordinates for the vertices of a cuboctahedron (of edge length $sqrt2$) centered at the origin are:
$$
beginarrayc
(pm1,pm1,0) \
(pm1,0,pm1) \
(0,pm1,pm1)
endarray
$$

which at least seems to confirm that I haven't misunderstood the construction.