This image is from John Meier's Introduction to the geometry of infinite groups on p. 47, where reflection groups are being introduced.
Now the claim is, that every edge in the tiling can be labelled either as $r,b$ or $g$ (which is also indicated by the different line styles in the picture).
This makes total sense. However, it is also being mentioned that $r, rgr$ and $rbr$ are the reflections in the sides of the triangle immediately above the gray triangle $\mathbb{T}$ in this tiling.
He explains that we can flip the triangle above down to $\mathbb{T}$ via $r$, then apply $g$ to reflect it to the left of $\mathbb{T}$ and after applying $r$ once again, we are in the desired position.
I'm still confused. After reflecting the triangle above $\mathbb{T}$, let's call it $\mathbb{T}'$, down to $\mathbb{T}$ via $r$, we have $r\mathbb{T}' = \mathbb{T}$. That's fine. Applying $g$ to reflect $r\mathbb{T}' = \mathbb{T}$ to the left is also fine, so we are now on the left side of $\mathbb{T}$, precisely at $rg\mathbb{T}'$.
But why can we reflect $rg\mathbb{T}'$ via $r$? The way I see it is that we can only reflect along the three "walls" (the three different line styles in each triangle) but not along the tip of the cone, can we?
The intuitive picture I am having in mind is this: Let's say we reflect the gray triangle $\mathbb{T}$ via $g$ to the left $g\mathbb{T}$. What I have in mind is that $g\mathbb{T}$ has the three walls:
$g$ is the "right wall", $r$ is the "left wall" and $b$ is the bottom wall of $g\mathbb{T}$.
In other words, if I were to apply $g$ and then $r$, I would move $\mathbb{T}$ as pictured here:
(sorry for the messy picture)
I hope I managed to express my confusion appropriately.
So my actual question: If the reflections on every single triangle are presicely $r,g,b$, then why is $rgr\mathbb{T}'$ the triangle on the left side of the triangle $\mathbb{T}'$ right above $\mathbb{T}$ rather than the left most triangle as pictured in my second picture?
Thanks for any help!
$r$, $g$, $b$ are 3 fixed lines of reflection, determined by the sides of the grey triangle in your first picture. Any point, hence any triangle, can be reflected across these fixed lines $r$, $g$, $b$. The meaning of "reflection across $r$" doesn't change based on which triangle you are reflecting. You are always reflecting across the same line.
Your last picture is incorrect: When you reflect $g\mathbb{T}$ by $r$, the reflection takes place over the original line $r$, the dark solid side of the original grey triangle $\mathbb{T}$. It does not mean to reflect $g\mathbb{T}$ across the dark solid side of $g\mathbb{T}$. After you perform the reflection of $g\mathbb{T}$ across $r$, you get the triangle directly above $g\mathbb{T}$. This is the same triangle as the reflection of $\mathbb{T}'$ across the side $g'$ (by which I mean the grey solid side of $\mathbb{T}'$), which is the point the author is making.