[example: Why One Coordinate Does Not Cover $S^1$]
Let $S^1=\{(x,y)\in\mathbb R^2:x^2+y^2=1\}$, and define $\gamma:(0,2\pi)\to S^1$ by $\gamma(t)=(\cos t,\sin t)$. Since $\cos$ and $\sin$ are smooth real functions, both coordinate functions of $\gamma$ are smooth. Also $\gamma(t)\in S^1$ because
\begin{align*}
\cos^2 t+\sin^2 t=1.
\end{align*}
The point $(1,0)$ is not in the image, since $\cos t=1$ and $\sin t=0$ force $t=0$ or $t=2\pi$ modulo $2\pi$, neither of which lies in $(0,2\pi)$. Conversely, every point of $S^1\setminus\{(1,0)\}$ has a unique polar angle $t\in(0,2\pi)$, so $\gamma$ parametrizes exactly the punctured circle.
This parametrization cannot be made into one coordinate chart on all of $S^1$ by merely adding an endpoint. If $0$ and $2\pi$ were both allowed, then
\begin{align*}
\gamma(0)=(\cos 0,\sin 0)=(1,0)
\end{align*}
and
\begin{align*}
\gamma(2\pi)=(\cos 2\pi,\sin 2\pi)=(1,0),
\end{align*}
so the map would not be one-to-one. If one instead keeps an open interval as the coordinate domain, then one endpoint of the interval is missing, and that missing endpoint is precisely the point where the circle closes.
The local-coordinate remedy is to use more than one chart. Let $N=(0,1)$ and $S=(0,-1)$, and define stereographic coordinates
\begin{align*}
\varphi_N(x,y)=\frac{x}{1-y}
\end{align*}
on $S^1\setminus\{N\}$ and
\begin{align*}
\varphi_S(x,y)=\frac{x}{1+y}
\end{align*}
on $S^1\setminus\{S\}$. For $t\in\mathbb R$, the inverse of $\varphi_N$ is
\begin{align*}
\varphi_N^{-1}(t)=\left(\frac{2t}{1+t^2},\frac{t^2-1}{1+t^2}\right),
\end{align*}
because
\begin{align*}
\left(\frac{2t}{1+t^2}\right)^2+\left(\frac{t^2-1}{1+t^2}\right)^2=\frac{4t^2+(t^2-1)^2}{(1+t^2)^2}=\frac{t^4+2t^2+1}{(1+t^2)^2}=1.
\end{align*}
On the overlap $S^1\setminus\{N,S\}$, the coordinate $t=\varphi_N(x,y)$ is nonzero, and the transition map is
\begin{align*}
(\varphi_S\circ\varphi_N^{-1})(t)=\frac{\frac{2t}{1+t^2}}{1+\frac{t^2-1}{1+t^2}}=\frac{\frac{2t}{1+t^2}}{\frac{2t^2}{1+t^2}}=\frac{1}{t}.
\end{align*}
The reverse transition is the same formula with the variables renamed, so both transition maps are smooth on $\mathbb R\setminus\{0\}$. Thus $S^1$ is not naturally described by one global interval coordinate; it is described by compatible local coordinates whose changes of coordinate are ordinary smooth functions.
[/example]