I have to calculate the inverse $f^{-1}(x)$ of $y=f(x)=2x-1$ and it is simple for this kinds of functions
Let $x=f(y)=2y-1$
$x+1=2y$
$\displaystyle\frac{x+1}{2}=y$
We now have the inverse $f^{-1}(x)=\displaystyle\frac{x+1}{2}$
But how am I goin to find the inverse of functions like $h(x)=f(2x)$?
Note that they are completely two different assignments
If what you need to invert is not given explicitly but contains some generic invertible function, you need to express the inverse using the inverse of that function.
You can think of $f(2x)$ as the composition of two functions. The function "multiplication by two", so $x \mapsto 2x$, and $f$. Let us call the multiplication by two function $m$, so $m(x)=2x$. Then $h(x)= (f \circ m)(x)$.
Presumably you heard something like $(f_1\circ f_2)^{-1} = f_2^{-1} \circ f_1^{-1} $.
So $h^{-1}(x) = (m^{-1} \circ f^{-1})(x)$ and you can determine $m^{-1}$ and $f^{-1}$ separately. The first should not be hard for you, and the second depends on $f$, but if $f$ is not given the best you can do is say it is $f^{-1}$, and your answer will involve $f^{-1}$.
More informally, you have $f(2x)$, if you want to undo that you have to first undo $f$, and then divide by two.