The framing obstruction lives in the normal bundle of the surgery sphere

Suppose $\nu$ is trivial and fix a smooth trivialization $\nu \cong S\times {\mathbb{R}}^q$. Apply [F2] in $\mathrm{int}M$; the compact sphere $S$ is closed there. The open tube domain contains the zero section, so finitely many product neighbourhoods give a common radius $r>0$ with $S\times D_r^q$ inside it. Restrict the tube to this closed disk bundle and rescale to obtain an embedding $S\times D^q\hookrightarrow \mathrm{int}M$. Its differential in the normal directions induces a framing (not necessarily the initially chosen trivialization, since [F2] specifies only its zero-section restriction). Thus $S$ is the underlying sphere of a framed embedded surgery sphere.

Conversely, the differential of $\phi$ at $(x,0)$ identifies $T_x{S}^p\oplus {\mathbb{R}}^q$ with $T_{{\phi }_0(x)}M$, and identifies the first summand with the tangent space of $S$. Passing to the quotient identifies the second summand smoothly with ${\nu }_x$, giving a trivialization $S\times {\mathbb{R}}^q\cong \nu$. Together with step 1.1 this proves (i), and applied to each embedded representative gives the equivalence of (a) and (b) in (iii).

For $q=1$ the normal bundle $\nu$ is a real line bundle over $S$, and $SO(1)$ is the trivial group, so orientability of $\nu$ means that its structure group reduces to the trivial group, i.e. that $\nu$ is trivial. By [F4] orientability is equivalent to $w_1(\nu )=0$, so $\nu$ is trivial exactly when $w_1(\nu )=0$, and by step 2.1 such a sphere admits no framing otherwise. This proves (ii).

In the below-middle range $p\le q-2$, assume additionally that $M$ is closed, and let $z\in {\pi }_p(M)$ be represented by a framed embedded surgery sphere $\phi$ as in (a). Then the hypothesis of [F8] is satisfied, and $z$ lies in the kernel of ${\pi }_p(M)\to {\pi }_p(M_{\phi })$: the datum of type (a) is what the construction uses to kill the class. A class whose embedded representatives all have nontrivial normal bundle has no representative of type (b) by step 2.1, hence none of type (a), so the construction does not apply to it; the converse implication is not claimed here. This proves (iii).

A framing of $\nu$ is a trivialization, hence an isomorphism $\nu \cong S\times {\mathbb{R}}^q$. Under this isomorphism the bundle carries $q$ everywhere linearly independent sections, in particular a nowhere-zero section, so if $\nu$ is oriented its Euler class vanishes by [F6], and all positive degree Stiefel-Whitney classes vanish by naturality [F7] because the trivial bundle is pulled back from a point. Therefore a nonzero Euler class, or a nonzero positive-degree Stiefel-Whitney class, of $\nu$ obstructs a framing; this is the line-bundle statement of (ii) in the case $q=1$. Vanishing of all these classes is necessary and is not claimed to be sufficient. This proves (iv).