What does $d \mathbb{P}(\omega)$ mean in expectation of r.v. $f$?
$$\mathbb{E} f = \int_{\Omega} f(\omega) d \mathbb{P}(\omega)$$
Yes sure it's some "infinitesimal", but should this mean that $\mathbb{P}(\omega)$ is a variable of $f$? Since in elementary integrals:
$$\int_X f(x) dx$$ and $x$ is a variable of $f$.
$\omega \in \Omega$ by def. but why is it $d \mathbb{P}(\omega)$ then and now just $d \omega$?
Part of what might be confusing to you already shows up in discrete probability. Here is an example: we roll a fair die, and let $X$ be the resulting number. We (that is, math teachers) say things like, the expected value of $X$ is $EX=\sum_{i=1}^6 i (1/6)$ and the expected value of $X^2$ is $EX^2=\sum_{i=1}^6 i^2 (1/6)$. Suppose we also roll an unfair die, and call the result $Y$, for which $P(Y=i) = p_i$ where not all the $p_i$ values are $1/6$. We say the expectations of $Y$ and $Y^2$ are $EY=\sum_{i=1}^6 i p_i$ and $EY^2 = \sum_{i=1}^6 i^2p_i$. In all cases the expectation is a weighted sum of the value we are talking about, and the system of weights we assign to the fundamental outcomes. In more advanced textbooks you will often see expressions like $EX=\int_\Omega X(\omega) P(d\omega)$, where $\omega$ plays the role of $i$ in my examples in this paragraph, where $P(\omega)$ plays the role of $p_i$ and so on.
In this way, we think of expectations as arising from two ingredients: a random quantity and the corresponding probability distribution. In your case, the expression $f(\omega)$ is the random quantity whose expectation we seek, and $d \mathbb P(\omega)$ is the corresponding probability distribution.
You might be also confused by the use of the letter $f$, which in textbooks is often used more-or-less exclusively as part of the probability law specification. Your example, however, is not an instance.