This is maybe not the answer you're looking for, but it's certainly too long for a comment.
First, there's no contradiction here, and nothing too strange. A 2-framing on $M$ is a framing of $TM\oplus TM$, which a priori doesn't have much to do with $TM$ (and moreover it comes with a splitting, which $TM$ doesn't).
Anyway, let me give an answer by analogy. Consider the case of $M = S^1$. Then $TM$ is trivial, but not canonically so: in this case a framing is the same thing as an orientation, or as a nowhere-vanishing section. $TM\oplus TM$ is also trivial, and either trivialisation of $TM$ induces a "diagonal" trivialisation of $TM \oplus TM$. (Note that Atiyah talks about the diagonal embedding $\Delta\colon SO(3) \to SO(3) \times SO(3)$, which then embeds in $SO(6)$ via an embedding $\iota$.) Moreover, the two induced trivialisations differ by multiplication by $-1$, which on a rank-2 bundle is isotopic to the identity. So the two trivialisations induced are equivalent (as trivialisations of a rank-2 bundle).
EDIT (What follows is slightly speculative.) This should boil down to showing that for every compact oriented 3-manifold $M$ and every map $\tau \colon M \to SO(3)$ (which measures a difference in trivialisations) the composition of maps $\iota\circ\Delta\circ\tau$ is homotopic to a constant map. This is non-trivial even in the case of $M = S^3$, since $\pi_3(SO(6)) = \mathbb{Z}$.
What mme (is saying that Aityah) is saying is that, for a 3-manifold $M$, the composition $\iota\circ\Delta\circ\tau$ gives an identification of the induced framings of $TM\oplus TM$ with $\mathbb{Z}$, and in particular there is a preferred framing (the one that corresponds to $0 \in \mathbb{Z}$). See mme's comments below for more details. (Thanks, mme!)