The distinction to be made is that a differentiable structure is a choice of maximal smooth atlas $\mathcal A$, but two different choices $\mathcal A$ and $\mathcal A'$ can lead to isomorphic smooth structures. As an example, the canonical smooth structure $\mathcal A$ on $\mathbb R$ that contains the smooth function ${\rm id}:\mathbb R\longrightarrow \mathbb R$ is isomorphic to the smooth structure $\mathcal A'$ that contains the smooth function $x\mapsto x^3$, although $\mathcal A'\neq \mathcal A$. Thus, although a manifold admits uncountably many different smooth structures, it may have finitely many isomorphism classes of such structures.
A sufficient condition for a codimension-$0$ manifold with non-empty manifold boundary embedded in $\mathbb{R}^n$ to have its manifold and topological boundary coincide may be that it is properly embedded.
Consider the following excerpt from Lee's Introduction to Smooth Manifolds (p. 120, 2nd edition):
If $M$ is a smooth manifold with or without boundary, a regular domain in $M$ is a properly embedded codimension-$0$ submanifold with boundary. Familiar examples are the closed upper half space $\mathbb{H}^n \subseteq \mathbb{R}^n$, the closed unit ball $\bar{\mathbb{B}}^n \subseteq \mathbb{R}^n$, and the closed upper hemisphere in $\mathbb{S}^n$.
Proposition 5.46 Suppose $M$ is a smooth manifold without boundary and $D \subseteq M$ is a regular domain. The topological interior and boundary of $D$ are equal to its manifold interior and domain, respectively.
The definition of proper embedded is found on p. 100 of the same:
An embedded submanifold $S \subseteq M$ is said to be properly embedded if the inclusion $S \hookrightarrow M$ is a proper map [preimages of compact sets are compact].
Since $\mathbb{R}^n$ is connected, a necessary condition for a submanifold with or without boundary of codimension $0$ to be properly embedded is that its manifold boundary is non-empty:
Proposition 5.49.(c) Suppose $M$ is a smooth manifold with or without boundary and $S \subseteq M$ is an embedded submanifold with boundary. Then $S$ is properly embedded if and only if it is a closed subset of $M$.
(See also this related question on Math.SE) This is because a codimension $0$ embedded submanifold with empty manifold boundary must be an open subset of $\mathbb{R}^n$, and if it is a proper subset, then it can not also be a closed subset of $\mathbb{R}^n$, since $\mathbb{R}^n$ is connected.
This doesn't say anything about my second question though, since Proposition 5.49(c) only applies to manifolds with boundary, and not proper embeddings of more general subsets.
Best Answer
I think Tu’s statement is fine:
A manifold, by definition, always has a dimension. Where are the charts going?
Usually when we say that a manifold “has boundary,” we mean that it has nonempty boundary.
After looking at some of Tu’s (non-standard!) definitions, I think you’re correct. An accurate statement might be