I am having difficulty with an argument in the proof below. So we assume $\tilde{S}$ to be an immersed submanifold with some topology other than the subspace topology and some smooth structure, and $S$ be the same set with the subspace topology and the smooth structure constructed from k-slice charts. Since $\tilde{S}$ is an immersed submanifold, the inclusion map $\tilde{i}: \tilde{S}\to M$ is an injective immersion and $\tilde{i}(\tilde{S})=S$. Because $S$ is embedded, Corollary 5.30 implies that $\tilde{i}$ as a map from $\tilde{S}$ to $S$ is also smooth.
But then I don't follow the composition below. How do we have $\tilde{i}=\tilde{i} \circ i$, where $i:S \to M$ is the inclusion map from the embedded manifold $S$ to $M$? Topologically this makes sense but in terms of manifolds, I can't quite grasp why we get the equality here because the right map is from the immersed submanifold to the ambient manifold $M$, and the left is a composition of a map from the immersed submanifold to the embedded submanifold and to the ambient manifold.
Or in a more general case (Theorem 5.33) where $S$ is a weakly embedded submanifold ($S$ may not have subspace topology), and $S'$ is another set with some topology and smooth structure with respect to which it is an immersed submanifold. Again, the proof uses the fact that $i':S' \to S$ is smooth since $S$ is weakly embedded, and follow the lines below that $i':S' \to M$ is a composition of $i':S' \to S$ and $i:S\to M$. How can we conclude that they are equal maps in terms of continuous and smooth maps?
I would greatly appreciate if anyone could clarify this to me.
I think you are getting confused by a common abuse of notation. Call $\tilde i:\tilde S\to M$ and $i:S\to M$ to the inclusions. And write $\tilde i':\tilde S\to S$ for the co-restriction, so $i\circ\tilde i'=\tilde i$. Since $i$ is an embedding, the co-restriction $\tilde i'$ is also smooth (see e.g. Corollary 5.30). Since $\tilde i$ is an immersion $d\tilde i_p=di_p\circ d\tilde i'_p$ is injective, then $d\tilde i'_p$ must be injective. The proof follows by noting that $\tilde i':\tilde S\to S$ is an immersion and a bijection, hence a diffeomorphism (Lee's Global Rank Thm).