There are a couple confusing things going on here.
First, basis covectors are often denoted something like $\mathrm dx^i$ which is visually similar to $dx^i$ (I'm writing them like so, differently, to emphasize these are different notions; it wouldn't surprise me to see $dx^i$ as a basis covector either though), so the relationship between a basis covector and a differential appears clear, even obvious.
And yet these things are not really even alike.
When you integrate a $k$-covector, what are you really doing? You're integrating on some (usually) $k$-dimensional manifold. If $e_1, e_2, \ldots$ are basis vectors of this manifold*, then an infinitesimal patch of this manifold is described using a $k$-vector $dV = (e_1 \wedge e_2 \wedge \ldots \wedge e_k) dx^1 dx^2 \ldots dx^k$.
What do $k$-covectors do? Well, usually we're told they eat $k$ vectors in all to give a scalar (or a scalar field, if in fact we have a $k$-covector field). Alternatively, they can be seen to eat a single $k$-vector.
So what you're really doing when you integrate a $k$-covector is this:
$$\int_U \omega \equiv \int_U \omega(e_1 \wedge e_2 \wedge \ldots \wedge e_k) \, dx^1 dx^2 \ldots dx^k$$
For some reason, people seldom even talk about the existence of $k$-vectors. Knowing they exist at all is really important. It turns the Riemann tensor into a map from $2$-vectors to $2$-vectors, for instance. Anyway, the object $\omega(e_1 \wedge \ldots \wedge e_k)$ is a scalar function and as such you clearly a classic Riemannian integral now.
*I don't denote that these basis vectors depend on the point of the manifold you take them at, but they do have this dependence.
The decomposition is not unique, because we can replace $\psi \wedge f^{\ast} \omega$ with $(f^{\ast}(g) \psi) \wedge f^{\ast} (g^{-1} \omega)$ where $g$ is some smooth function $Y \to \mathbb{R}_{>0}$. Of course, this doesn't effect the pushforward, since $\int_{f^{-1}(y)} f^{\ast}(g) \psi = g(y) \int_{f^{-1}(y)} f^{\ast}(g)$ and then $\left( \int_{f^{-1}(y)} f^{\ast}(g) \psi ) (g^{-1} \omega) \right)= g(y) g(y)^{-1} \omega = \omega$. More generally, if $\phi = \alpha \wedge f^{\ast}(\beta \wedge \gamma \wedge \delta)$ with $\alpha$ an $r-k$ form, $\beta$ and $\gamma$ $k$-forms and $\delta$ a $p-r-k$ form, then $\phi = (\alpha \wedge f^{\ast} \beta) \wedge f^{\ast}(\gamma \wedge \delta) = (-1)^k (\alpha \wedge f^{\ast} \gamma) \wedge f^{\ast}(\beta \wedge \delta)$ are two such factorizations.
I think the simplest way to prove indepedence of the decomposition is to describe the pushforward in terms of its integrals. For $\phi$ any compactly supported $p$-form on $X$, and $B$ any embedding of a closed $(p-r)$-dimensional ball into $Y$, we have $\int_B f_{\ast} \phi = \int_{f^{-1}(B)} \phi$ (using Fubini). Since a $(p-r)$-form is defined by its integrals over all $(p-r)$-dimensional balls, this proves uniqueness of such a $f_{\ast} \phi$ (but not existence, use Bott and Tu's argument for that.)
Best Answer
I can see why you might be confused, but the hypotheses ($\omega$ is compactly supported and an orientation form) imply that $M$ must be compact. So that's the only case to which that statement applies.