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 $\sigma\circ (d^0\times d^1).$

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 $t:[2]\times [1]\to K$ 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 $d^1:\ulcorner\to [2]\times [1],$ 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 $\sigma\circ (d^0\times d^1).$

Note that if we rotated the $[2]\times [1]$-diagram by $90^\circ$ to be vertical, the induced map would be $\Sigma f$: in particular, this proves that there's no way to get the triangle-generating rectangles for $f$ and $g$ both into a connected planar diagram without flipping one of their orientations.

This follows from the following claims. Let me know if the last two seem to need further fleshing out. For me this stuff requires visualizing the zeroes in suspension squares as upward- or downward-pointing cones on the object being suspended.

• 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 $(0,0)^*.$

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 $\sigma\circ (d^0\times d^1).$