Proof of $\det(\textbf{ST})=\det(\textbf{S})\det(\textbf{T})$ in Penrose graphical notation For two matrices $\textbf{S}$ and $\textbf{T}$, a proof of $\det(\textbf{ST})=\det(\textbf{S})\det(\textbf{T})$ is given below in the diagrammatic tensor notation.

Here $\det$ denotes the determinant.
Why can the antisymmetrizing bar be inserted in the middle because "there is already antisymmetry in the index lines"?
For an introduction to the notation, you can refer to Figures 12.17 and 12.18 below.


 A: (see explanatory note if confused)
Penrose writes "The antisymmetrizing bar can be
inserted in the middle term because there is already antisymmetry in the index
lines that it crosses." in the caption for Figure 13.8
What he means is that "antisymmetrization" is an idempotent tensor operator. This, in turn, follows from the fact that the antisymmetrizer can be viewed as a projection onto the totally antisymmetric subspace, and the fact that projection operators are idempotent. (See Antisymmetrizer)
Recall that an operator $I$ is idempotent iff $I^2=I$. In other words, we can replace the single bar in the diagram with two bars. That is all.

Explanatory Note: I have not posted images because (a) they are under copyright and (b) they are not relevant to the crux of the question.

I looked at the google cache of this webpage and noticed that all the attached diagrams were screenshots from the book "The Road to Reality" by Roger Penrose, which I have access to.
Figures 12.17 and 12.18 are located on pages 241 and 242, respectively.
The proof KalEl is referring to is found as Figure 13.8, on page 261.
A: I think Penrose made a mistake on this page and forgot a factorial.   He defines the antisymmetrization bar in graphical notation as not including a factorial, so it is not idempotent.    The middle step should pick up an extra factor of $\frac{1}{n!}$.   You can tell it needs it because the right hand side is supposed to be the product of two determinants, which according to (a) each get a factor of $\frac{1}{n!}$.
The reason you can add the additional antisymmetrization bar in the middle is the same reason why the second step needs to pick up an extra factorial.    If you expand out the bar, each summand gets a $\pm1$ depending on the permutation.   But since the triangles are all the same, and the Levi-Civita symbol on the bottom is also antisymmetric, you can undo all the permutations and all the signs and they all add up to $n!$.
