The $-\Sigma f$ bit follows from the following claims. Let me know if the last two seem to need further fleshing out. Keeping track of the differences in things that are all called $0$ is the trickiest part of the calculus of stable derivators.
- When you restrict $j_!i_*(f)$ along $d^1\times d^2,$ you get $(0,0)_*X.$
- When you restrict $j_!i_*(f)$ along $d^0\times d^1,$ you get $\sigma^*(0,0)_*Y,$ where $\sigma$ is the nontrivial automorphism of $\ulcorner.$
- By definition of the cogroup structure on suspended objects, a map $(0,0)_*X\to \sigma^*(0,0)_*Y$ evaluating to $f$ under $(0,0)^*$ induces $-\Sigma f$ when extended to the suspensions.
- Since the natural transformation $\alpha:d^1\times d^2\Rightarrow d^0\times d^1$ (pretty sure Groth has the variance backwards here) has $(0,0)\to (1,0)$ as its component at $(0,0),$ the component of $\alpha^*$ at $j_!i_*(f)$ evaluates to $f$ under $(0,0)^*.$
As for the formal reflection (not rotation!) of the rectangle on the right, observe that the rectangle defining the distinguished triangle generated by $g$ arises by restricting $j_!i_*(f)$ along the functor $t:[1]\times [2]\to K$ sending $(0,0)\mapsto (1,0),(0,1)\mapsto (2,0),(1,0)\mapsto (1,1),$ and so on. In particular, the composition of $t$ with $d^1:\ulcorner\to [1]\times [2],$ which is the standard identification of $(2,2)^*j_!i_*(f)$ with $\Sigma Y,$ sends $(1,0)$ to $(1,2)$ and $(0,1)$ to $(2,0).$ That is, the standard identification is equal to $\sigma\circ (d^0\times d^1).$