In the wikipedia link provided by the OP, the probability integral transform in the univariate case is given as follows
Suppose that a random variable $X$ has a continuous distribution for which the cumulative distribution function(CDF) is $F_X$. Then the random variable $Y=F_X(X)$ has a uniform distribution.
PROOF
Given any random variable $X$, define $Y = F_X (X)$. Then:
$$ \begin{align} F_Y (y) &= \operatorname{Prob}(Y\leq y) \\
&= \operatorname{Prob}(F_X (X)\leq y) \\
&= \operatorname{Prob}(X\leq F^{-1}_X (y)) \\
&= F_X (F^{-1}_X (y)) \\
&= y \end{align} $$
$F_Y$ is just the CDF of a $\mathrm{Uniform}(0,1)$ random variable.
Thus, $Y$ has a uniform distribution on the interval $[0, 1]$.
The problem with the above is that it is not made clear what the symbol $F_X^{-1}$ represents. If it represented the "usual" inverse (that exists only for bijections), then the above proof would hold only for continuous and strictly increasing CDFs. But this is not the case, since for any CDF we work with the quantile function (which is essentially a generalized inverse),
$$F_Z^{-1}(t) \equiv \inf \{z : F_Z(z) \geq t \}, \;\;t\in (0,1)$$
Under this definition the wikipedia series of equalities continue to hold, for continuous CDFs. The critical equality is
$$\operatorname{Prob}(X\leq F^{-1}_{X} (y)) = \operatorname{Prob}(X\leq \inf \{x : F_X(x) \geq y \})= \operatorname{Prob}(F_X (X)\leq y)$$
which holds because we are examining a continuous CDF. This in practice means that its graph is continuous (and without vertical parts, since it is a function and not a correspondence). In turn, these imply that the infimum (the value of inf{...}), denote it $x(y)$, will always be such that $F_X(x(y)) = y$.
The rest is immediate.
Regarding CDFs of discrete (or mixed) distributions, it is not (cannot be) true that $Y=F_X(X)$ follows a uniform $U(0,1)$, but it is still true that the random variable $Z=F_{X}^{-1}(U)$ has distribution function $F_X$ (so the inverse transform sampling can still be used). A proof can be found in Shorack, G. R. (2000). Probability for statisticians. ch.7.
There is an error in your calculation (it seems you would get $F_Y(0) = 1/3$, but this probability should be $0$). Note that (for $y\geq 0$), $P(X^2\leq y)$ is actually equal to $P(-\sqrt{y}\le X \le \sqrt{y}) = \color{red}{F_X(\sqrt{y}) - F_X(-\sqrt{y})}$. Try and compute this red part for $y\in [0,4]$. You will need to remember that $F_X(t)$ will become $0$ if $t < -1$.
Best Answer
What your textbook says is actually not entirely true. If two random variables have the same cumulative distribution function, then their density functions are equal almost everywhere. For instance, a perfectly valid density function for a random variable uniformly distributed over the open interval $(0, 1)$ is a function which equals one over $(0, 1)$ and zero otherwise. Or we could just as easily define the density to equal one at the endpoints if we wanted. Being even more extreme we could have it take on completely arbitrary values on any countable subset of $\mathbb{R}$, the only requirement is that the density functions need to integrate to the same value over sets of the form $(-\infty, a]$. If this is the case then it's clear we'll arrive at the same distribution function.
In general though we take the density to be the derivative of the cumulative distribution function wherever the latter is differentiable and just set it to zero everywhere else.