Change of variable formula and probability distributions

Question

I think I have a misconception that I hope someone can help me clarify, I do not have a background in measure theory so that might be the problem.

Say we have two random variables $X$ and $Y$ distributed according to $f_X(x)$ and $g_Y(y)$ respectively, without necessarily a common support. Both random variables are related by some monotonic function $h(.)$ such that $Y = h(X)$. And we have the following, which I think is called equivalence in distribution but please correct me if not: \begin{align} P(Y \leq m) &= P(X \leq h^{-1}(m)) \\ \int_{A}g_Y(y)dy &= \int_{h^{-1}(A)}f_X(x)dx \end{align}

According to the change of variable formula, the density of $Y$ is related to the density of $X$ by the following relation: \begin{align} g_Y(y) = f_{X}(h^{-1}(y)).|d[h^{-1}(y)]/dy| \end{align}

I am wondering because there are certain situations where we have a closed form of both $g_Y$ and $f_X$ and computing the above probabilities does give the same answer, but writing the dependence between both as in the change of variable formula gives a different form for $g_Y$ that depends on the inverse of $g$. For example, a very simple case where the distribution of the CDF of a random variable is known to have the uniform distribution: \begin{align} Y = F_X(X) \end{align}

$F_X$ is the CDF of $X$ which is a valid monotonic transformation. We know that $Y \sim U[0,1]$, but if you apply the change of variable formula we get something that is the density of $f_X$ which is a function of the INVERSE CDF multiplied by the derivative of that inverse, which does not usually have a closed form. If I put this result for $g_Y(y)$ in the above integral is something cancelling out like the $dy$? How does the derivative of the inverse of $g$ disappear in the above equality of the integrals?

Excuse me in advance if these questions are very basic, I am a self-learner and I do not have any colleagues who might be able to help with these kinds of silly questions.