The lasso KKT conditions from Part 1 can be reproduced in a different variable: one nonnegative scale $\gamma_j$ per coefficient. This auxiliary system separates activation from sign and gives the MM construction its natural coordinates.
Introduce $\v\gamma=(\gamma_1,\ldots,\gamma_p)^\top\in\bb{R}_+^p$ and define
$$ \m\Gamma:=\operatorname{diag}(\sqrt{\gamma_1},\ldots,\sqrt{\gamma_p}), $$ $$ \begin{aligned} \m G(\v\gamma) &:=(\m I+\m X\m\Gamma^2\m X^\top)^{-1}\\ &=\left(\m I+\sum_{j=1}^p\gamma_j\v x_j\v x_j^\top\right)^{-1}. \end{aligned} $$Recover a coefficient vector from $\v\gamma$ by
$$ \begin{aligned} \v\beta(\v\gamma) &:=\m\Gamma^2\m X^\top\m G(\v\gamma)\v y,\\ \beta_j(\v\gamma) &=\gamma_j\v x_j^\top\m G(\v\gamma)\v y. \end{aligned} $$The identity $(\m I+\m X\m\Gamma^2\m X^\top)\m G=\m I$ then gives
$$ \v r(\v\gamma) :=\v y-\m X\v\beta(\v\gamma) =\m G(\v\gamma)\v y. $$Thus the same quantity controls both the coefficient and its residual correlation. Write
$$ \tau_j(\v\gamma):= |\v x_j^\top\m G(\v\gamma)\v y| =|\v x_j^\top\v r(\v\gamma)|. $$Consider the convex objective on the nonnegative orthant
$$ \begin{aligned} g(\v\gamma) &:=\v y^\top\m G(\v\gamma)\v y +\alpha^2\v 1^\top\v\gamma,\\ &\hspace{6em}\v\gamma\in\bb{R}_+^p. \end{aligned} $$Because $\partial\m G/\partial\gamma_j=-\m G\v x_j\v x_j^\top\m G$, its coordinate derivative has the particularly simple form
$$ \begin{aligned} \frac{\partial g}{\partial\gamma_j} &=\alpha^2-(\v x_j^\top\m G(\v\gamma)\v y)^2\\ &=\alpha^2-\tau_j(\v\gamma)^2. \end{aligned} $$The constraint $\gamma_j\ge0$ requires a zero derivative in the interior and a nonnegative one at the boundary. Since $\alpha$ and $\tau_j$ are nonnegative, the KKT conditions reduce to
An active scale sits exactly on the auxiliary KKT boundary.
An excluded scale may lie anywhere below the boundary.
It is useful to name the signed auxiliary KKT gap $\xi_j(\v\gamma):=\alpha-\tau_j(\v\gamma)$. A positive scale requires $\xi_j=0$; a zero scale is valid when $\xi_j\ge0$ and violating when $\xi_j<0$.
If $\gamma_j>0$, then $\beta_j=\gamma_j\v x_j^\top\v r$ has the same sign as the residual correlation. The equality $\tau_j=\alpha$ therefore becomes $\v x_j^\top\v r=\alpha\operatorname{sign}(\beta_j)$. If $\gamma_j=0$, then $\beta_j=0$ and $\tau_j\le\alpha$ becomes $|\v x_j^\top\v r|\le\alpha$. These are exactly the two lasso KKT cases from Part 1.