It is a standard result that on a domain $\Omega$ in $\mathbb{R}^n$, $T=\sum_{|I|=q}T_{I}dx_{I}$ for any $q$-current $T$, where $T_{I}$ is a $0$ current or a distribution on $\Omega$ for each multi-index $I$. Now, for an $r$-current $T$ and an $s$-form $\alpha$ on $\Omega$, define an $r+s$-current as $(T\wedge\alpha)(\beta)=T(\alpha\wedge\beta)$ for any $n-r-s$-form $\beta$ on $\Omega$. Then, in the above we have $T=\sum_{|I|=q}T_{I}\wedge dx_{I}$ on $\Omega$. At this stage, if we start with a general smooth manifold $M$(orientable) and a $q$-current $T$ on $M$, can we write $T=\sum_{\alpha}T_{\alpha}\wedge\omega_{\alpha}$ on $M~?$ (Here $T_{\alpha}$ and $\omega_{\alpha}$ are distributions and $q$-forms on $M$ for each $\alpha$).
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The answer is mostly yes, although you may need to be careful with non-compact manifolds.
First look at the case of a compact manifold. Cover $M$ with a finite number of coordinate charts $\phi_i: U_i \to M$ and take a subordinate partition $\psi_i$ of $1$, so that $\sum_i \psi_i = 1$ and $\mathop{supp} \psi_i \subset \subset\phi_i(U_i)$. Now on each chart you can use the pushforward to consider $S_i := (\phi_i^{-1})_* (T \wedge \psi_i)$ as a current with compact support on $U_i \subset \mathbb{R}^n$. So apply your standard result on $S_i$ to get $S_i = \sum_I S_{i,I} dx_I$ and push this back to $M$, in order to get $$T \wedge \psi_i = \sum_I \underbrace{(\phi_i)_* S_{i,I}}_{T_{i,I}} \underbrace{(\phi_i)_* dx_I}_{\omega_{i,I}}$$ Sum over $i$ to get the result. (You may need to cut of the $\omega_{i,I}$ outside of $\mathop{supp} \psi_i$ to guarantee smoothness, but I'll leave that to you)
For a non-compact manifold you can only choose a locally finite cover, so either your sum will be infinite or you will have to combine non-overlapping coordinate charts somehow, I vaguely remember that this can be done.
You can probably also try to take a different route, I think the main condition is finding $\omega_a$ such that they span the space of all $q$-forms in each point. If you have those than you can probably proceed similarly to the original proof.
