Defining manifold coordinates without an embedding

The issue here is that a circle is most naturally a subspace of $\mathbb{R}^n$, usually $\mathbb{R}^2$.

But we can choose other ways to think about a circle that don't describe it as an embedded submanifold of $\mathbb{R}^n$. For example, $[0,1]$ with the points $0$ and $1$ identified is a circle. The quotient $\mathbb{R}/\mathbb{Z}$ is also a circle. Each description of a circle has its uses, but we usually choose the embedding into $\mathbb{R}^2$ for visualization purposes.

The moral is that some manifolds in some contexts most naturally present as an embedded submanifold of $\mathbb{R}^n$, but they also exist as abstract manifolds (a topological space plus smoothly compatible charts).

Perhaps an example might help.

Consider the definition of projective space $\mathbb R P^n$, which is the quotient space of $\mathbb R^{n+1} - \{0\}$ under the equivalence relation $p\sim q$ if there exists $r \ne 0$ such that $p=rq$. This space is not given to us by an embedding in any Euclidean space.

Nonetheless, we can work out coordinate charts for this space, and verify that the overlap maps are smooth, and so this is a manifold. These coordinate charts are as follows.

Let $[p] \in \mathbb RP^n$ be the equivalence class of $p = (p_0,...,p_n) \in \mathbb R^{n+1} - \{0\}$.

For $i=0,...,n$ let $U_i = \{[p] \mid p_i \ne 0\} \subset \mathbb R P^n$. Then the function $\phi_i : U_i \to \mathbb R^n$ defined by $\phi_i[p] = \frac{1}{|p_i|}(p_0,...,p_{i-1},p_{i+1},...,p_n)$ is well-defined, and it is a coordinate chart for $\mathbb R P^n$. Taking these all together we get an atlas of coordinate charts $\{\phi_i : U_i \to \mathbb R^n \mid 0 \le i \le n\}$.

I think the problem here is that we are so used to visualizing the circle as a geometric object that we consider it (intuitively, and our intuition is correct) embedded in $\mathbb R^2.$ But, if we simply define $M=\{(x,y):x^2+y^2=1\}$ where we give $M$ the subspace topology of $\mathbb R^2$, then, with the charts as in your post, $M$ becomes a bonafide manifold in the abstract sense. The fact that we already know inuitively that $M$ is locally homeomorphic to $\mathbb R$ is something that the abstract definition of $M$ is designed to make rigorous. And the fact that $M$ is a $\textit{subset}$ of $\mathbb R^2$ is incidental. An example of an abstract manifold for which geometric intuition at least for me, does not work, wold be a Lie group, $\mathrm{GL}_n(\mathbb{R})$ for example.