I'm afraid I'm not familiar with your first construction, but a very close one is the Alexander complex/Alexander-Spanier cohomology, where we replace your first order jets along the diagonals by all jets along the diagonal (functions on the formal neighborhood). This complex and its identification with de Rham cohomology is explained beautifully in Appendix A of Constantin Teleman's "Borel-Weil-Bott Theory on the moduli of G-bundles over a curve", available eg. <a href="http://math.berkeley.edu/~teleman/paperlist.html">here</a>. Presumably there's a simple relation between this and your construction 1, ie everything reduces to 1-jets, but I haven't thought about this.

One succinct way to describe this Alexander construction is as derived functions on the de Rham stack of M - the quotient of M by the formal neighborhood of the diagonal.
Writing this stack as a simplicial space we get a cosimplicial algebra of functions on it, which is quasiisomorphic to the de Rham complex.

On the other hand, your second construction is a description of the odd tangent bundle, which is a version of the derived loop space of a manifold. In other words, your second construction is a rephrasing of the description of de Rham cohomology as periodic cyclic homology - or more precisely, your $R^{0,1}$ action is the Connes differential on the Hochschild homology of functions on $M$. So I would consider the equivalence of the two approaches the fact that functions on the de Rham space and cyclic homology are two well-known approaches to describing de Rham cohomology.

This approach to de Rham cohomology via loop spaces is the subject eg of my paper with Nadler <a href="http://arxiv.org/abs/1002.3636">Loop Spaces and Connections</a>. 
In fact we introduce there a space we call the unipotent loop space of a (derived) stack, which is maps $Map(A^1[1],M)$ from the shifted version of the affine line just as you define (keeping track of Z-gradings here). For a manifold or general scheme this agrees with the shifted tangent bundle (or rather tangent complex if you're singular) $T_M[-1]$, but this is no longer true for a stack -- for example for $BG$ you get unipotent elements in $G$, up to conjugation. Rather its formal completion is the formal completion of the odd tangent bundle (or if you prefer its relative tangent space is the odd tangent bundle). In any case you can describe de Rham cohomology (or more generally the whole category of D-modules) very succinctly in terms of this mapping space and its translation action by $A^1[1]$ as you mention (this is a linearized form of the circle action on the loop space of $M$). 

The relation between the two constructions can be described as a case of Koszul duality -- specifically the groupoid algebra of the de Rham stack is the algebra of differential operators on $M$, which is Koszul dual to the de Rham complex as a dga. (Again this is described in some detail in the above paper - the idea goes back at least to a paper of Kapranov.) This Koszul duality gives an equivalence between sheaves equivariant for the de Rham/Alexander groupoid and dg modules for the de Rham complex. This relation between the two constructions is extremely general - certainly true for arbitrary smooth schemes and stacks, but in fact doesn't require smoothness if you interpret everything correctly. The claim is that the Koszul duality between dg Lie algebras and dg commutative algebras (in an algebroid setting) exchanges the tangent complex (or the algebra of differential operators) and the de Rham complex (or if you prefer, the de Rham complex is the Chevalley-Eilenberg complex of the tangent complex).

In any case, functions on the de Rham stack (construction 1) always calculate the de Rham cohomology of a space (or derived stack or whatever), while functions on maps from the odd line with the translation action as differential gives this answer as long as you're not a stack (i.e. periodic cyclic homology of a stack remembers some things you'd like to forget maybe, namely the cohomomlogy of inertia, once you throw that away the two answers always agree).