About the construction of the restriction of a Vector Bundle in John M. Lee's book
(I am not 100% sure, but here's what I believe what the author really meant.)
Short Answer: If $\pi:E\to M$ is a (topological) vector bundle and $S\subseteq M$ is any subspace, then we denote by $E|_S\to S$ the restriction of $\pi:E\to M$ as you have defined. However, when $\pi:E\to M$ is a smooth vector bundle and $S\subseteq M$ is an immersed submanifold, we give $E|_S$ a smooth structure that makes it into an immersed (but possibly not embedded) submanifold of $E$. There is a good reason for this: Even though $E|_S$ may not be embedded in $E$, it has a nice property.
An example of the case where $E|_S$ is not embedded: Let $E=M=\mathbb{R}^2$, and let $\pi=\operatorname{Id}_{\mathbb{R}^2}$. Clearly $\pi:E\to M$ is a smooth vector bundle over $M$ (of rank $0$). Now let $S$ be the figure eight, which is an immersed submanifold in $M$. The chart lemma makes $\pi^{-1}(S)=E|_S$ into an immersed submanifold of $E$. Since $S$ has a self intersection, we cannot make $E|_S$ into a smooth manifold (nor even a topological manifold!) with respect to the subspace topology inherited from $E$. Thus we see that the chart lemma endowed $E|_S$ with a topology different from the subspace topology, i.e., $E|_S$ is not embedded.
Long Answer: Suppose $\pi :E\to M$ is a smooth vector bundle, and let $S$ be an immersed submanifold in $M$. The chart lemma (Lemma 10.6 in Lee's book) shows that $E|_S=\pi^{-1}(S)$ has a (unique) topology and a smooth structure with respect to which the restriction $\pi_S:E|_S\to S$ of $S$ is a smooth vector bundle with the following property:
- If $\Phi:\pi^{-1}(U)\to U\times \mathbb{R}^k$ is a smooth local trivialization of $E$ over an open set $U\subseteq M$, then its restriction $\Phi|_S:\pi^{-1}(U\cap S)\to(U\cap S)\times \mathbb{R}^k$ is a smooth local trivialization of $E|_S$.
It is easy to check that $E|_S$ is an immersed submanifold of $E$.It is also easy to check that in the case when $S$ has the subspace topology inherited from $M$ and is an immersed submanifold of $M$ (i.e., when $S$ is embedded), the topology of $E|_S$ given by the chart lemma agrees with the subspace topology inherited from $E$, so there is no conflict in what the symbol $E|_S$ means in this case. If $S$ is merely immersed, the chart lemma may endow $E|_S$ with a topology different from the subspace topology, so there is a little bit of ambiguity in what the symbol $E_S$ means. However, when $S$ is immersed but not embedded, we are keenly aware of the fact that it is is not "part of $M$", in the sense that it does not have the subspace topology. So there is little chance of confusion.
Although it may not be embedded in $E$, it resembles an embedded submanifold in the following sense:
If $\sigma:S\to E$ is a smooth map such that $\pi\circ\sigma=\operatorname{Id}_S$, then $\sigma:S\to E|_S$ is a smooth section. (You can replace "smooth" by "continuous" here.)
Before we get to the proof, let me note that this property explains justifies the arguments in Example 10.10 and lemma 10.35. Now we get back to the proof.
(Proof.) Let $p$ be any point in $S$. Let $\Phi:\pi^{-1}(U)\to U\times\mathbb{R}^k$ be any smooth local trivialization of $E$ with $p\in U$. By hypotheis, the map $$\Phi\circ\sigma:U\cap S\to U\times\mathbb{R}^k$$ is smooth. This, along with the hypothesis that $\pi\circ\sigma=\operatorname{Id}_S$, implies that there are smooth functions $\sigma^i:U\cap S\to \mathbb{R}$ such that $\Phi\circ\sigma(q)=(q,\sigma^1(q),\dots,\sigma^k(q))$ for $q\in U$. Now $\Phi|_S\circ\sigma:\pi^{-1}(U\cap S)\to(U\cap S)\times\mathbb{R}^k$ is given by exactly the same formula: $$\Phi|_S\circ\sigma(q)=(q,\sigma^1(q),\dots,\sigma^k(q)).$$ The smoothness of $(\sigma^i)$ implies that $\Phi|_S\circ\sigma$ is smooth, so the restriction of $\sigma$ to $U\cap S$ is smooth. Since $p$ was arbitrary, the claim follows.$\square$