Given a differentiable function $F(x)$ the differential $d(F(x))$ can be defined according to the rule
$$\int f(x) dF(x) = \int f(x) F'(x) dx$$
or in other words $dF(x) = F'(x) dx$. Intuition for this comes from thinking of $dF$ as the small perturbation of $F(x)$ as we perturb $x$ by $dx$:
$dF = \frac{dF}{dx} dx = F'(x) dx$.
Now suppose we are interested in $dG$ for $G(x) = F(x)^2$. The integral rule says
$$\int f(x) dG(x) = \int f(x) G'(x) dx = 2\int f(x) F(x)F'(x) dx =2\int f(x) F(x)\frac{dF}{dx} dx=2\int f(x) F(x) dF $$
and so $dG(x) = 2F(x) dF(x)$ or $dF(x) = \frac{dF^2(x)}{2F(x)}$
HOWEVER!
Thinking of $dF$ as a probability distribution instead leads to the following derivation:
$$dF(x) = P(\omega : x \le F(\omega) \le x + dx ) $$
$$=P(\omega : x^2 \le F^2(\omega) \le (x + dx)^2 ) $$
$$=P(\omega : x^2 \le G(\omega) \le (x + dx)^2 ) $$
$$=P(\omega : x^2 \le G(\omega) \le x^2 + 2xdx+ (dx)^2 )$$
$$=P(\omega : x^2 \le G(\omega) \le x^2 + 2xdx )$$
$$ = 2x P(\omega : x^2 \le G(\omega) \le x^2 + dx ) $$
$$ = 2x dG(x^2) $$
Hence we get $dF(x) = 2x dG(x^2)$ and so $dG(x^2) = \frac{dF(x)}{2x}$. Replace $x$ with $\sqrt x$ to get $dG(x) = \frac{dF(\sqrt x)}{2 \sqrt x}$.
In particular if $dG$ is uniform ($dx = dG(x)$) the first formula gives $dx = 2F(x) dF(x)$ and $dF(x) = \frac{dx}{2F(x)}$. In order to get $dG(x) = dx$ we need $G'(x) = 1$. Hence $G(x) = x$ and $F = \sqrt x$. So we get $dF(x) = \frac{dx}{2\sqrt x}$.
However!
If $dG$ is uniform we also have $dG(x^2) = dx$ and the second formula gives $dx = dG(x^2) = \frac{dF}{2x}$ and so $dF = 2x dx$.
These derivations cannot both be right. Which one is correct and where is the mistake?
The problem is you are squaring the distribution but also interpreting as if you squared the random variable with the distribution.
Let $X_F$ be the random variable corresponding to distribution $F(x)$.
Let $Y_F$ be the random variable corresponding to distribution $F^2(x)$.
Then : $$dF(x) = P(\{\omega : x \leq X_F(\omega) \leq x + dx \})$$
It is not: $$dF(x) = P(\{\omega : x \leq F(\omega) \leq x + dx \})$$
Infact: $$dF(x) = \mu(\{y : F(x) \leq y \leq F(x + dx) \})$$ $$dF(x) = \mu(\{y : F(x)^2 \leq y^2 \leq F(x + dx)^2 \})$$ $$dF(x) = \mu(\{y : F(x)^2 \leq y^2 \leq (F(x)+dF(x))^2 \})$$ $$dF(x) = \mu(\{y : F(x)^2 \leq y^2 \leq F(x)^2+2F(x)dF(x) \})$$ $$dF(x) = \mu(\{y : G(x) \leq y^2 \leq G(x)+2F(x)dF(x) \})$$ $$dF(x) = \mu(\{y : G(x) \leq y^2 \leq G(x+dx) \})$$