Why do we have a conservation of probability during the transformation? I don't know what it is called in the literature but whenever we have a transformation of random variables (x -> y) with the probability distributions of $g(y)\ 
\&\ f(x)$, we write saying that probability is equal, the following:
$$ f(x)dx = g(y)dy$$
Now, y and x are related through a function, let's take it to be $J(x) = y$. The next step is generally using the delta function for identifying the new probabilities $P(y)$ through $$ P(y) = \int dx\ \delta( y-J(x) ) f(x)  $$
Hence my two-part question:
How, starting from this, I can derive the first equation logically. The general derivation treats derivatives as fractions which I don't consider to be correct.
In some of the books, it is given directly. So, how does it make sense intuitively, too?
Comment
Since it is pointed out, the complete expression for P(y) would be

$$ g(y)dy = P(y) = \int dx\ \delta( y-J(x) ) f(x)  $$

 A: Your $P(y)$ is in reality $g(y)$ and the $dy$ in your last expression does not make sense. Here is an informal way to obtain $g$:  Let the random variable $X$ have density $f$ and suppose that $Y=J \circ X$ has density $g$.
Then given an interval (or Borel set) $A$ we have by definition of the densities that $$\int_A g(y) \; dy = P(Y\in A) = P(J\circ X \in A) = P(X\in J^{-1}(A))= \int_{J^{-1}(A)} f(x)dx$$
You may here introduce a factor of one, formally through a $\delta$-"function" integral and write the RHS as:
$$ \int_{J^{-1}(A)} \left( \int_{\Bbb R} \delta(y-J(x)) dy \right) f(x) \; dx =
\int_{J^{-1}(A)} \left( \int_{A} \delta(y-J(x)) dy \right) f(x) \; dx
.$$
Note that for every $x\in J^{-1}(A)$ we have $J(x)\in A$ so the inner integral equals one even when restricting the $y$-integral to $A$. Now, if you don't worry too much about Fubini when delta-functions are involved you get by interchanging integrals and then enlarging (without changing the values) the inner domain:
$$ \int_{A} \left( \int_{J^{-1}A } \delta(y-J(x)) f(x) \; dx \right) \; dy
= \int_{A} \left( \int_{{\Bbb R}} \delta(y-J(x)) f(x) \; dx \right)  \; dy
=: \int_A g(y) \; dy$$
thus identifying $g(y)$ with $\int_{\Bbb R} \delta(y-J(x)) f(x) dx$.
Assuming $J$ to be $C^1$, you can make this rigorous and get a formula for $g$
iff $J$ is regular almost surely on the values taken by $X$,
i.e. iff:
$$ P\left(X\in \{x\in {\Bbb R}: J'(x)\neq 0 \}\right)=1.$$
In a mathematical proof I would, however, rather avoid the delta function approach and use change of variables directly. But as I am not sure what you are after I stop here.
