The method in his slides differs from his original paper. First I will explain Milnor's original construction of exotic $7$-spheres, from his article
- On manifolds homeomorphic to the $7$-sphere, Annals of Mathematics, Vol. 64, No. 2, September 1956.
I'll answer your question about his construction mentioned in the slides after that.
First, he defines a smooth invariant $\lambda(M^7)$ for closed, oriented $7$-manifolds $M^7$. To do this, we first note that every $7$-manifold bounds an $8$-manifold, so pick an $8$-manifold $B^8$ bounded by $M^7$. Let $\mu \in H_7(M^7)$ be the orientation class for $M^7$ and pick an "orientation" $\nu \in H_8(B^8,M^7)$, i.e. a class satisfying
$$\partial \nu = \mu.$$
Define a quadratic form
$$H^4(B^8,M^7)/\mathrm{Tors} \longrightarrow \mathbb{Z},$$
$$\alpha \mapsto \langle \alpha \smile \alpha, \nu \rangle,$$
and let $\sigma(B^8)$ be the signature of this form. Milnor assumes that $M^7$ has
$$H^3(M^7) \cong H^4(M^7) \cong 0,$$
so that
$$i: H^4(B^8,M^7) \longrightarrow H^4(B^8)$$
is an isomorphism. Hence the number
$$q(B^8) = \langle i^\ast p_1(B^8) \cup i^\ast p_1(B^8), \nu \rangle$$
is well-defined. Then Milnor's $7$-manifold invariant is
$$\lambda(M^7) \equiv 2q(B^8) - \sigma(B^8) \pmod 7.$$
Milnor shows that $\lambda(M^7)$ does not depend on the choice of $8$-manifold $B^8$ bounded by $M^7$.
Now let us turn to the specific $7$- and $8$-manifolds that Milnor considers. As you noted, he looks at the total spaces of $S^3$-bundles over $S^4$. The total space of such a bundle is a $7$-manifold bounding the total space of the associated disk bundle. $S^3$-bundles over $S^4$ (with structure group $\mathrm{SO}(4)$) are classified by elements of
$$\pi_3(\mathrm{SO}(4)) \cong \mathbb{Z} \oplus \mathbb{Z}.$$
An explicit isomorphism identifies the pair $(h,j) \in \mathbb{Z} \oplus \mathbb{Z}$ with the $S^3$-bundle over $S^4$ with transition function
$$f_{hj}: S^3 \longrightarrow \mathrm{SO}(4),$$
$$f_{hj}(u) \cdot v = u^h v u^j$$
on the equatorial $S^3$, where here we consider $u \in S^3$ and $v \in \mathbb{R}^4$ as quaternions, i.e. the expression $u^h v u^j$ is understood as quaternion multiplication.
Let $\xi_{hj}$ be the $S^3$ bundle on $S^4$ corresponding to $(h,j) \in \mathbb{Z} \oplus \mathbb{Z}$. For each odd integer $k$, let $M^7_k$ be the total space of the bundle $\xi_{hj}$, where
\begin{align*}
h + j & = 1, \\
h - j & = k.
\end{align*}
Milnor shows that
$$\lambda(M^7_k) \equiv k^2 - 1 \pmod 7.$$
Furthermore, he shows that $M^7_k$ admits a Morse function with exactly $2$ critical points, and hence is homeomorphic to $S^7$. Clearly we have
$$\lambda(S^7) \equiv 0,$$
so if
$$k \not\equiv \pm 1 \pmod 7,$$
then $M^7_k$ is homeomorphic but not diffeomorphic to $S^7$, and hence is an exotic sphere. In particular, $S^7$, $M^7_3$, $M^7_5$, and $M^7_7$ are all homeomorphic to one another but all pairwise non-diffeomorphic.
Now, in the slides, the space $E$ should be the total space of the disk bundle associated to an $S^3$ bundle $\xi_{hj}$ over $S^4$ with
\begin{align*}
h + j & = 1, \\
h - j & = k
\end{align*}
for some odd integer $k$, as described above. Then $E$ is an $8$-manifold with boundary $\partial E$ homeomorphic to $S^7$. Now, if $\partial E$ is diffeomorphic to $S^7$, then we can glue $D^8$ to $E$ along their common boundary via a diffeomorphism
$$f: \partial E \longrightarrow S^7$$
in order to get a smooth manifold
$$E' = E \cup_f D^8.$$
If $f$ is not a diffeomorphism, then $E'$ is not necessarily smooth. So in showing that
$$p_2(E') \notin \mathbb{Z},$$
Milnor proves by contradiction that no such diffeomorphism $f$ can exist, since Pontrjagin numbers of a manifold are integers. So in that case $\partial E$ would be homeomorphic to $S^7$ but not diffeomorphic, and hence an exotic $7$-sphere.
Best Answer
Ideally you'd ask whoever wrote those lecture notes; it's a bit remiss of them not to include a picture or something.
Here's my guess, this feels like the most "obvious" cobordism $S^1 \to [0, 1]$: start with a cylinder (the identity cobordism $S^1 \to S^1$) and pinch one end shut. (So it's kinda like a zipped-up purse, I guess.)