On page 56 of Mao's SDE book, the author considers an existence result for SDE under the local Lipschitz assumption:
Here $a\bigvee b:=\max\{a,b\}$. What he does is to reduce the argument to the global Lipschitz case by considering
$$ f_{n}(x, t)=\left\{\begin{array}{ll} f(x, t) & \text { if }|x| \leq n \\ f(n x /|x|, t) & \text { if }|x|>n, \end{array}\right. $$ also the same for $g$. It is claimed that $f_n$ and $g_n$ satisfies the global Lipschitz and linear growth condition, but I do not see how.
Let us omit $t$. For $|x|\leq n$ and $|y|\leq n$, this is clear, and for $|x|>n$ and $|y|>n$, we have \begin{align} |f_n(x)-f_n(y)|&=|f(nx/|x|)-f(ny/|y|)|\leq \sqrt{K_n} |nx/|x|-ny/|y||\\ &=\sqrt{K_n}\, n|x/|x|-y/|y||\leq 2n\, \sqrt{K_n} |x-y|, \end{align} where in the last inequality, $|x|,|y|\geq 1\implies |x/|x|-y/|y||\leq 2|x-y|$ is used; cf. $\left|\frac{x}{|x|}-\frac{y}{|y|}\right|\leq |x-y|$, for $|x|, |y|\geq 1$?.
But I do not see how to show the global Lipschitz continuity for other cases; when, WOLG, we have $|x|\leq n$ and $|y|>n$.
A related question is Cutting off local Lipschitz gives global Lipschitz, which is on another way of cooking up a function like this.
If you have points on a line, then you can apply the triangle identity to get $$ |f_n(x_M)-f_n(x_0)|\le\sum_{k=0}^{M-1}|f_n(x_{k+1})-f_n(x_k)|\le L\sum_{k=0}^{M-1}| x_{k+1} - x_k |=L|x_m-x_0| $$ if all the parts satisfy the Lipschitz condition. Thus you do not need to consider mixed cases as you can take the inner points on the sphere $|x_k|=n$ so that the segments are either completely outside or completely inside.
Note that the link for the outside inequaltity was for $|x|\land|y|>1$. Here you have $|x|\land|y|>n$, so that the inequality scales as $$ |\hat y-\hat x|=\frac{\Bigl||x|\,y-|y|\,x\Bigr|}{|x|\,|y|} \le\frac{|x|\,|y-x|+\Bigl|(|y|-|x|)\,x\Bigr|}{|x|\,|y|} =2\frac{|y-x|}{|y|}\le\frac2n|y-x|. $$
This above inequality is true for all vector norms. Specifically for the Euclidean norm one can use Euclidean geometry to get $$ |\hat y-\hat x|=2d = \frac2{|x|+|y|}\left( |x| d + |y|d\right)\le\frac{2}{|x|+|y|}(|y-z|+|z-x|)=\frac{2}{|x|+|y|}(|y-x|) $$ where $z$ is the intersection point of the angle bisector of the rays through $x$ and $y$ and the segment $[x,y]$, see the sketch below
Along these lines more directly $$ |\hat y-\hat x| =\frac{\Bigl|y-x+|x|\hat y-|y|\hat x\Bigr|}{|x|+|y|} \le\frac{\Bigl|y-x\Bigr|+\Bigl||x|\hat y-|y|\hat x\Bigr|}{|x|+|y|} =\frac{2|y-x|}{|x|+|y|} $$ as $|x|\hat y-|y|\hat x$ is the reflection of $y-x$ on the angle bisector between $x$ and $y$.