Let me make sure I understand you correctly. Given a compact Riemannian manifold of dimension $n$, any smooth $k$-chain induces a functional on the space of $k$-forms, and any $n-k$-form induces a functional on the space of $k$-forms. You're asking what the relationship is between these two functionals.
Answer: There is no overlap between these two kinds of functionals. Why? Suppose a $k$-chain $\sigma$ induces the same functional as a $n-k$-form $\omega$. Clearly, $\omega$ must be nonzero on some open set $U$ that doesn't intersect any part of $\sigma$. Then if $\alpha$ is a test $k$-form, then $\sigma(\alpha)$ doesn't depend on the values of $\alpha$ on $U$, but $\omega(\alpha)$ certainly does.
Now, if you look at these functionals at the level of homology/cohomology, then as you probably know, the relationship is Poincare duality. That is, $[\sigma]$ and $[\omega]$ induces the same functionals on the $k$-th cohomology iff they are Poincare dual.
Here's what currents have to do with the story. $k$-currents are essentially defined to be all linear functionals on the space of smooth $k$-forms (with appropriate topology). Then the two types of functionals you describe, the $\sigma$'s and the $\omega$'s are now specific examples of $k$-currents. If you know much about distributions, then you pretty much already know about currents--you just define everything via integration by parts. The boundary operator on currents generalizes both the boundary operator on chains and the $d^*$ operator on forms. Not surprisingly, the homology of currents gives you the (real) homology of $M$. The reason I immediately said that the $\sigma$'s and $\omega$'s have no overlap is that, in analogy with plain old distributions, the $\omega$'s are like smooth functions, while the $\sigma$'s are like singular measures supported on sets of measure zero.
I don't know much history, but I think that your idea was what lead de Rham to invent currents.
Let $E$ denote a vector bundle over a manifold $M$ equipped with a metric, and $L_p(E)$ the space of measurable sections of $E$ with finite $L_p$ norm.
Obviously, in general, one can't identify $L_p(E)$
with an $L_p$ space of vector valued functions.
First assume that $M$ is compact. To understand $L_p(E)$, we use a finite set of trivializations $(U_i, h_i)$ of $E$ which cover $M$. Each trivialization identifies $E|_{U_i}$ with $U_i \times\mathbf R^n$ (or $U_i \times\mathbf C^n$). We choose the trivializations such that the bundle norm is equivalent to the euclidean norm, i.e. bounded from above and below. Then an $L_p$ section of $E$ is equivalent to a set of $\mathbf R^n$ (or $\mathbf C^n$)-valued measurable functions ${f_i}$ on $U_i$ satisfying the transition law, such that $\sum \|f_i\|_p$ is finite.
Using this one can easily extend all the basic theorems to $L_p(E)$. In particular, one shows that $L_p(E)$ is equivalent to the completion of $C^{\infty}(E)$ (the space of smooth
sections on $M$) w.r.t. the $L_p$ norm.
If $M$ is noncompact, we write it as the union of locally finite compact subsets $A_i$, such that the intersections of $A_i$ have zero measure. Then the $L_p$ norm of a section
$s$ is given by
$(\sum \int_{A_i} |s|^p)^{1/p}$.
Then the arguments for the compact case can easily be carried over. (We argue on each
$A_i$, and then combine.)
Best Answer
Differential $k$-forms can be integrated along a submersion with $d$-dimensional fibers, which yields a differential $(k-d)$-form.
Fiberwise integration (alias pushforward) of differential forms is a standard operation, described in many expository texts. A fairly detailed presentation is given in Chapter VII of