Personally I do not like the concept of an immersed submanifold as presented by Lee and other authors. The situation is this:
We have an injective immersion $j : P \to M$. This induces a unique topology and a unique smooth structure on the image set $S = j(P)$ such that $\bar j : P \stackrel{j}{\to} S^*$ becomes a diffeomorphism, where $S^*$ denotes the set $S$ endowed with the appropriate topology and smooth structure. This $S^*$ is called an immersed submanifold of $M$.
The problem here is that $j$ is in general no homeomorphism from $P$ onto its image $j(P)$ with the subspace topology inherited from $M$. Thus, although we call $S^*$ an immersed submanifold, it is in general not even a topological subspace of $M$. In my opinion this is somewhat confusing, but of course it is a matter of taste.
However, even if $S^*$ should not be a topological subspace of $M$ globally, it is always locally one. This means that each $q \in S^*$ has an open neigborhood $V$ in $S^*$ which is an emdedded submanifold of $M$. Note that $V$ is open in $S^*$, but in general not open in $S$ with the subspace topology inherited from $M$.
In my opinion a better formulation of Lee's Theorem 5.29 would be this:
Let $f :N \to P$ be a continuous function from a smooth manifold $N$ to a smooth manifold $P$ and let $j : P \to M$ be an immersion (which is not required to be injective). If $F = j \circ f$ is smooth, then $f$ is smooth.
Lee's theorem is a corollary. We are given an immersed submanifold $S^*$ of $M$ and a smooth $F : N \to M$ such that $F(N) \subset S^*$ and $\bar F : N \stackrel{F}{\to} S^*$ is continuous. This $S^*$ is associated to an injective immersion $j : P \to M$ such that $\bar j : P \stackrel{j}{\to} S^*$ is a diffeomorphism. The function $f = (\bar j)^{-1} \circ \bar F : N \to P$ is continuous and $j \circ f = F$ is smooth, thus $f$ is smooth. Hence $\bar F = \bar j \circ f$ is smooth.
Our above theorem can be proved exactly as Lee's Theorem 5.29. Let us now see why $(V_0, \widetilde{\psi})$ is a smooth chart for $S^*$.
Since $V$ is open in $S^*$ and $V$ is an embedded submanifold of $M$, it suffices to show that $(V_0, \widetilde{\psi})$ is a smooth chart for $V$. But this follows from a theorem in Lee's book which says that each subspace $E \subset M$ which is an embedded submanifold has a unique smooth structure such that the inclusion map $E \to M$ is a smooth embedding. A smooth atlas on $E$ generating this smooth structure is given precisely by the collection of all $(W_0 = W \cap E, \pi \circ \psi \mid_{W_0})$ where the $(W,\psi)$ are smooth charts for $M$ which are slice charts for $E$ centered at the points $q \in E$.
Perhaps it is somewhat confusing that the smooth structure on $V$ is primarily induced by $j_V : j^{-1}(V) \stackrel{j}{\to} V$. However, since we know that $V$ is an embedded submanifold of $M$, this smooth structure agrees with the above smooth structure induced by slice charts.
The lemma is not true for immersed submanifolds. A counterexample is the irrational winding of the torus (see Example 4.20 in Lee). This is the image of the map
$$
\gamma:\mathbb{R}\rightarrow\mathbb{T}^{2}:t\mapsto (e^{2\pi it},e^{2\pi i\alpha t}),
$$
where $\alpha$ is an irrational number. It is an immersed, but not an embedded submanifold.
Since $\gamma(\mathbb{R})$ is dense in $\mathbb{T}^{2}$, the zero function is the only function that vanishes on $\gamma(\mathbb{R})$. So for any vector field $Y\in\mathfrak{X}(\mathbb{T}^{2})$, we have $f|_{\gamma(\mathbb{R})}=0 \Rightarrow Y(f)|_{\gamma(\mathbb{R})}=0$. But not all vector fields on $\mathbb{T}^{2}$ are tangent to $\gamma(\mathbb{R})$.
Best Answer
To make the solution to this question clear,I will explain with example:
Just consider the $S$ be figure eight curve in $M = \Bbb{R}^2$.$E$ over $M$ just take rank-0 vector bundle $E= \Bbb{R}^2$.where $\pi$ associated to it is identity map. $E|_S$ as a set is just figure eight curve.
Let's see if we only use the construction same as embedded submanifold what will happen for $E_S$
that is construct the local trivialization as $\Phi_U:\pi_{S}^{-1}(U\cap S) \to (U\cap S) \times \{0\}$.We can see the local trivialization here is just all the identity $\Phi_U:U\cap S \to U\cap S$.
Recall we use this $\varphi\circ\Phi_U$ as a local chart for $E|_S$ in slice chart lemma. And $U\cap S$ should be the form of local chart for $S$.But $S$ is immersion $U\cap S$ is not local chart for $S$(near origin)
In order to solve this problem,we need to consider to shrink it into some locally embedded neiborhood.as explained in the link.