Suppose I embed a manifold-with-boundary $M$ in some $\mathbb{R}^n$. Are there conditions (necessary, sufficient, or both) that can help determine when the topological boundary of $M$ is equal to the manifold boundary?
By "topological boundary," I'm referring to $\text{Bd } M$, which is the closure minus the interior (relative to $\mathbb{R}^n$).
By "manifold boundary," I mean the boundary $\partial M$ that is specified in the definition of "manifold-with-boundary."
If you embed a smooth $m$-manifold $M$ smoothly in $\mathbb{R}^n$ then the "topological boundary" of $M$ is the closure of $M$. As locally each point of $M$ has a neighbourhood in $\mathbb{R}^n$ where $M$ looks like $\mathbb{R}^m$, then no point of $M$ is interior.
When $m=n$ and $M$ is compact then, yes (at least in the smooth case) the topological and manifold boundaries coincide.
I expect the above hold for topological embeddings but won't swear to it; they need not be locally flat (see nasties like the Alexander horned sphere),
In order to complement and clarify Chapman's answer:
For a topological manifold $M$, possibly with boundary, I will use the notation $\partial M$ to denote its boundary and $int(M)=M\setminus \partial M$ its interior (both are understood here in the sense of manifold topology).
For a subset $Y$ of a topological space $X$, I will use the notation $Int(Y)$ to denote its topological interior, i.e. the union of all open subsets of $X$ contained in $Y$. Accordingly, I let $Fr(Y)$ denote the frontier of $Y$ in $X$, which is $cl(Y)\setminus Int(Y)$. (I prefer the terminology "frontier" to that of "topological boundary," which appears in the OP.)
Suppose now that $M$ is an $m$-dimensional topological manifold, possibly with boundary, $m\le n$, and $f: M\to E^n$ is a topological embedding. Then the following proposition relates the boundary $M$ and the frontier of $f(M)$ in $E^n$:
Proposition. 1. Suppose that $m< n$. Then:
(a) $f(M)\subset Fr(f(M))$.
(b) If $M$ is compact, then $f(M)= Fr(f(M))$.
(a) $f(int(M))= Int(f(M))$ and $f(\partial M)\subset Fr(f(M))$.
(b) If $M$ is compact, then $f(\partial M)= Fr(f(M))$. If $M$ is noncompact, this equality might fail.
Proof. 1a is an immediate consequence of the invariance of domain theorem: An open subset of $E^m$ cannot be homeomorphic to an open subset of $E^n$, implying that $Int(f(M))$ is empty.
1b. Since $M$ is compact, $f(M)$ is closed in $E^n$, hence, $Fr(f(M))= f(M)\setminus Int(f(M))= f(M)$.
2a. $f(int(M))\subset Int(f(M))$ is a consequence of the invariance of domain theorem again. Suppose that $y=f(x)\in Int(f(M))$. Then there is an open ball neighborhood $V$ of $y$ contained in $R^n$, By applying the invariance of domain theorem to $f^{-1}: f(M)\to M$, we see that $x$ is an interior point of $M$. Thus, $f(int(M))= Int(f(M))$. In other words, $f(\partial M)\cap Int(f(M))=\emptyset$. It follows that $f(\partial M)\subset Fr(f(M))$.
2b. Since $M$ is compact, $f(M)$ is closed in $E^n$, hence, $Fr(f(M))=f(M)\setminus Int(f(M))= f(M)\setminus f(int(M))= f(\partial M)$.
To get an example of a noncompact manifold where the equality fails, take, for instance $M=[0,1)$ and $f$ the identity inclusion $M\to {\mathbb R}=E^1$. Then $Fr(M)= \{0, 1\}$, while $\partial M= \{0\}$. Similar examples exist in all dimensions. qed
