@NicholasKuhn It depends. In algebraic topology we usually (but not always) mean stably framed and shorten it to "framed". But, for example, in the cobordism hypothesis one talks of the framed bordism category and in this case one means unstable tangential framings. So it all depends on context. Perhaps you would prefer the term "parallelizable"? I will change the title.

@archipelago That is a nice idea. I think you are suggesting something along the following lines. Suppose that X is a stably framed manifold where the obstruction to destabilizing is non-trivial. Suppose for concreteness that [X] generates a $\mathbb{Z}/3\mathbb{Z}$ in $\pi^s_*$. Then the obstruction will vanish on 2[X], which also generates this group. But 2[X] is just two disjoint copies of X, so how could it all of a sudden become framable? Really the obstruction lives in a $\mathbb{Z}/2$ for each component of the manifold in question.

@archipelago Yes. The 7-sphere has a "Lie group" framing coming from viewing it as the unit octionians. It is not really a Lie group since it is not associative, but there is enough structure to talk about "left-invariant vector fields" and this induces a trivialization of the tangent bundle like for Lie groups. I believe that with this framing it represents $\sigma$.

@PanagiotisKonstantis I don't think so. As you observe the Lie group framing on $S^1$ gives the non-trivial element of $\Omega^{fr}_1$. The disjoint union of two circles, each with the Lie group framing gives a tangentially framed manifold representing the trivial element of $\Omega^{fr}_1$. Also for the sake of this question the empty manifold should also probably be considered a tangentially framed manifold.

No, the question is whether the class is represented by a manifold with trivial tangent bundle, not trivial normal bundle.

@DylanWilson No, in general the core will correspond an $E_\infty$-space but it is usually not grouplike, so is not an infinite loop space (in general). Your proposal does work when the core is grouplike, but that is a strong condition. It is equivalent to saying every fully dualizable object is actually invertible.

You modify the proof as follows. Without assuming orientability, then there are two cases. (1) The neighborhood of the loop gamma is a cylinder (which is the case described in Andy's note) or (2) it is a Mobius strip. In case (2) you cut it out and glue in a single disk. This still gives a surface with larger Euler characteristic (increased by 1 this time), so is still covered by induction.