Example of a situation when $P(X<Y)$ is not equal to $P(X^{2} < Y^{2})$?

Let $X$ be constantly $-1$ and $Y$ be constantly $0$. Then $P(X<Y) = 1 $ and $P(X^2 <Y^2) = 0$

Non-degenerate r.v.'s, taking the same set of values:

Let $X$ be $0$ with probability $1/3$ and $-1$ with probability $2/3$; and $Y$ (independent of $X$) be $0$ with probability $2/3$ and $-1$ with probability $1/3$.

$X,Y$ are discrete, non-degenerate, and both take values in $\{-1,0\}$. But $$ \Pr[X<Y] = \frac{2}{9}, \qquad \Pr[X^2<Y^2] = \frac{1}{9} $$

Previous answer (with a non-degenerate r.v., not the same domain):

Let $X$ be uniform on $\{-1,1\}$, and $Y$ be uniform on $\{0,1\}$. Then $$ \Pr[X<Y] = \Pr[X=-1] = \frac{1}{2} $$ but since $X^2=1$ with probability one and $Y^2=Y$, $$ \Pr[X^2<Y^2] = \Pr[Y>1] = 0\,. $$

You can cook up many such examples by letting $X$ taking negative values (but with $Y\geq 0$ a.s.).