In my particular example, I have a volume form on a one-dimensional manifold, so i.e. this volume form $dx$ is closed. Now, I was wondering, does the integral function $f(y):=\int_{x_0}^{y} dx$ define $f$ locally as a function?
Since I got the impression in the comments that my question is somewhat unclear. Why does it hold for a general closed form on a manifold $M$ that the integral defines $f(y):=\int_{x_0}^{y} dx$ locally $f$ as a function?
If $\omega$ is a closed $1$-form on a smooth manifold $M$, then in any contractible open set $U$, you can define a function $f$ such that $df=\omega$. This is a special case of a the Poincaré lemma, which says that every closed form is locally exact.
To define $f$, choose a point $x_0\in U$ and let $f(x) = \int_{\gamma_x}\omega$, where for each $x\in U$, $\gamma_x$ is some specific path from $x_0$ to $x$. Because closedness of $\omega$ and contractibility of $U$ guarantee that the integral is path-independent, this is sometimes written as $\int_{x_0}^x\omega$.
For more details, see Chapter 11 of my Introduction to Smooth Manifolds (2nd ed.).