This is an interesting problem, thanks for sharing! It looks like a correct solution is available at the link you gave, but I'll provide an alternate one here as well. (I'll leave the "animated sales videos" link for now as it just links back here.)

An alternative solution is as follows. Label the cube $ABCD-EFGH$ (meaning the base is square $ABCD$ and square $EFGH$ is above $ABCD$, so $\overline{AE}$, $\overline{BF}$, etc. are all edges). We are looking for the number of paths (without loss of generality) from $A$ to $G$ along the surface of the cube. Since the paths all have length $9$, they are the shortest paths possible. Therefore any such path will really take place in $2$ of the faces of the cube.

Consider first paths that travel in the faces $ABFE$ and $BCGF$. "Unfolding" these two faces we get the diagram shown below:

In this diagram, the number of paths from $A$ to $G$ is $\displaystyle \binom{9}{3}$ as we need to place $3$ 'ups' and $6$ 'rights' in $9$ spots.

There are $3$ faces adjacent to $A$, and for each of these faces there are $2$ choices for which other face would be included in paths. For each of these $3\cdot 2 = 6$ pairs of faces, there are $\displaystyle \binom{9}{3}$ paths.

HOWEVER, we are overcounting. For example, the path directly following $ABFG$ could be considered to be in the pair of faces listed above, or in $ABFE$ and $EFGH$, or $ABCD$ and $BCFG$, etc.

For example, any path that contains the point $F$ must continue directly to $G$ and thus would be counted as part the pair $ABFE$ and $BCGF$ as well as the pair $ABFE$ and $EFGH$. Using similar methods to above, there are $\displaystyle \binom{6}{3}$ paths from $A$ to $F$ that we are counting twice.

Similarly there are $\displaystyle \binom{6}{3}$ paths we are overcounting that contain the point $C$, and $\displaystyle \binom{6}{3}$ paths we are overcounting that contain the point $H$. This same reasoning gives us we are also overcounting $\displaystyle \binom{6}{3}$ paths that contain the points $B$, $D$, or $E$.

Putting this all together there are $$6\cdot \binom{9}{3} - 6\cdot \binom{6}{3} = 384$$ paths in total.

Let us know if you have any questions. If anyone else has alternative solutions, please share them as well!