Two very different behaviors show up depending on whether you invert 6 or not. Over Z[1/6] (and hence over Z(p) for p>3) the situation is completely rigid and all vector bundles split as sums of powers of the Hodge line bundle ω. At p=3 (and similarly at p=2) there are many nontrivial extensions and indeed indecomposables of all ranks.

Here are precise statements and proofs/sketches that address your two strategies.

1. Over Z[1/6] (in particular over Z(p) for p>3): every vector bundle splits as a sum of line bundles

There is a convenient presentation of the moduli stack of elliptic curves after inverting 6. Let MF := Z[1/6][c4,c6] with the usual grading by weight (deg c4 = 4, deg c6 = 6), and let Δ = (c4^3 − c6^2)/1728 of weight 12. The stack of elliptic curves (with a section) is M_{1,1}[1/6] ≅ [Spec MF[Δ^{-1}] / G_m], where G_m acts via the grading. (Equivalently, this is the stack of elliptic curves with a trivialization of ω, modulo the G_m that rescales that trivialization.) This is standard; see for example Deligne–Rapoport, Katz–Mazur, or Fulton–Olsson.

For a linearly reductive group scheme G (and G_m is linearly reductive over any base), quasi-coherent sheaves on [Spec R/G] are equivalent to G-equivariant quasi-coherent sheaves on Spec R; for vector bundles this says: Vect(M_{1,1}[1/6]) ≅ finite projective Z-graded MF[Δ^{-1}]-modules.

Now MF[Δ^{-1}] is a localization of a polynomial ring in two variables over Z[1/6]. By Quillen–Suslin (and stability under localization), every finitely generated projective module over MF[Δ^{-1}] is free. A graded finite projective module is therefore a finite direct sum of graded shifts of the free module R: M ≅ ⊕i R⟨d_i⟩. Under the above equivalence, the shift by d corresponds to tensoring with ω^⊗d. Thus every vector bundle on M{1,1}[1/6] is a finite direct sum of powers of ω. In particular Ext^1 between line bundles vanishes, in agreement with the calculations coming from the Adams–Novikov E_2-term at primes p>3.

This completely answers the classification over Z[1/6] (and after any base change to which 6 is invertible): there are no indecomposables beyond line bundles, and ω^⊗12 is trivial (as in Fulton–Olsson).

2. At p = 3: nontrivial extensions and indecomposable vector bundles

When 3 is not inverted, one cannot reduce Weierstrass equations to the short form y^2 = x^3 − c4 x − c6, so the simple presentation by a G_m-quotient of an affine scheme breaks down; instead one uses the full Weierstrass Hopf algebroid, whose structure group has a unipotent part in characteristic 3. This is exactly what shows up in computations such as Tilman Bauer’s: various Ext-groups between line bundles become nonzero, and there are genuine indecomposable vector bundles.

Your rank-2 example is the universal de Rham bundle: for the universal elliptic curve π: E → M_{1,1} the Hodge filtration gives an exact sequence 0 → ω → H^1_dR(E/M_{1,1}) → ω^{-1} → 0 whose extension class lies in Ext^1(ω^{-1}, ω) ≅ H^1(ω^2).

Over Z[1/6] the group H^1(ω^2) vanishes (so the sequence splits), but over Z_(3) it does not. One elementary way to see the nontriviality is to restrict to the residual gerbe at the point j = 0 in characteristic 3. There the stabilizer contains μ_3 (which is nonreduced, hence not linearly reductive in characteristic 3). The vector bundle H^1_dR of the supersingular elliptic curve at j = 0 carries a μ_3-representation; the Hodge filtration identifies a 1-dimensional μ_3-subrepresentation (ω) and the 1-dimensional quotient (ω^{-1}). In characteristic 3 this representation is not semisimple: μ_3 acts unipotently on H^1_dR with a single Jordan block of size 2 (equivalently, the Cartier operator is nilpotent). Therefore the sequence does not split as a μ_3-representation, hence cannot split as a vector bundle on the stack. Consequently, the extension class is nonzero, and your candidate is indeed an indecomposable rank-2 bundle on M_{1,1} over Z_(3).

One can go a bit further:

• The group Ext^1(ω^{-1}, ω) over F_3 is nonzero (in fact, one finds a Z/3 in many presentations; this matches Bauer’s calculation for the Weierstrass Hopf algebroid at p=3), and the above de Rham extension class generates it.

• More generally, because the residual gerbe at j = 0 contains Bμ_3 and representations of μ_3 in characteristic 3 are not semisimple, one obtains indecomposable vector bundles of arbitrary rank on M_{1,1}⊗F_3 by taking iterated self-extensions and then extending them from the gerbe to the whole moduli stack. In particular, a Birkhoff–Grothendieck type splitting theorem fails at p = 3.

Comments on your two strategies

• Your first strategy is essentially formal over Z[1/6]: the quotient presentation by G_m and Quillen–Suslin give you the complete splitting into line bundles. Over Z_(3), the appearance of nontrivial Ext^1 groups is not an artifact of the calculation; they measure the genuine failure of semisimplicity caused by non–linearly-reductive stabilizers in characteristic 3.

• Your second strategy gives a correct and natural example. The nontriviality can be certified by restriction to the residual gerbe at the supersingular point j = 0 as above. One can also identify the extension class with the Kodaira–Spencer/weight-2 “E2” class in Katz’s theory of modular forms; modulo 3 this class becomes genuinely modular (since p−1 | 2), and it does not vanish, which is another way to see that the extension is nonzero.

Summary

• Over any base where 6 is invertible, every vector bundle on M_{1,1} is a direct sum of powers of ω.

• At p = 3 (and similarly at p = 2) there are many indecomposable vector bundles. The H^1_dR extension 0 → ω → H^1_dR → ω^{-1} → 0 is a basic nontrivial example; its nontriviality can be seen by restriction to the residual gerbe at j = 0 in characteristic 3. In this regime, a clean classification akin to the line bundle case is not available; the category already contains the non-semisimple representation theory of μ_3, leading to families of indecomposables of arbitrary rank.