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.
On any Riemannian manifold, any differential operator has a formal adjoint.
This has nothing to do with functional analysis, it is a pure calculus fact that uses little more than Stokes theorem.
This is because taking adjoints is linear, preserves (or rather reverses)
composition and commutes with multiplication by real-valued functions.
It is also clear how the adjoint of an order 0 operator (i.e. vector bundle homomorphisms) has to be formed.
The last and most important step is to construct the adjoint of a vector field on functions. Once this is done,
the existence of an adjoint is proven, because the above operators generate the algebra of differential operator. This last argument also generalizes
to
differential operators on vector bundles. The formula below shows you also - for free - that the adjoint is a differential operator, which is
something you have to work on if you define adjoints on the (pre) Hilbert-space level.
Let $X$ be a vector field, viewed as a differential operator on compactly supported function. To compute the adjoint of $X$, start with the formula
$\int_M Lie_X \alpha = \int_M d \iota_X \alpha + \int_M \iota_x d \alpha =0$ for any $n$-form $\alpha$ and any vector field $X$. The first equality is the Cartan infinitesimal homotopy formula. The second summand is
zero for degree reasons, the first by Stokes. Now let $\omega$ the volume form and $f,g$ two functions. Therefore
$0= \int_M Lie (fg\omega)= \int_M (Xf) g \omega + \int_M f f (Xg) \omega+ \int_M fg \cdot div (X) \omega$ by the definition of the divergence.
Hence the adjoint of $X$ is $-X - div (X)$.
So: Once you know how define the volume form on a semi-Riemannian manifold, you know that formal adjoints exists. So you can say that
taking adjoints is a canonical operation on the algebra of differential operators, depending only on the volume form.
The argument above can be easily made into a formula for the adjoint once your operator is given in local coordinates. Whether this formula is useful or enlightening is a matter of taste.
For the exterior derivative, that formula can be written in terms of the Hodge star. This is a (rather simple) theorem. I do not know whether there is any other formula for $d^*$.
Pseudodifferntial operators also have adjoint, which is proven in a different fashion.
It is more difficult to show that the formal adjoint agrees with the Hilbert space adjoint (and I am not really qualified to say much on it and refer you to a book or a real expert)
Best Answer
Fixing a $k$ for simplicity, there are many inner products on $\bigwedge^k (M)$ (which I would usually denote $\Omega^k (M)$). Since $\bigwedge^k T^* M$ is a vector bundle, there are, for example, Sobolev $H^s$ inner products on its space of smooth sections for any natural number $s$. See, for example, Palais, Foundations of Global Non-Linear Analysis.