Inspired by this question, $\mathbb{R}$ can be given the atlas $\mathcal{A} = \{f(x) = x : x \in \mathbb{R}\}$ or the atlas $\mathcal{B} = \{g(x) = x^3 : x \in \mathbb{R}\}$.
These are incompatible because there exist combinations of $f,g$ and their inverses which produce non smooth functions $\mathbb{R} \rightarrow \mathbb{R}$, namely $f \circ g^{-1} = \sqrt[3]{x}$, so this fails to be a diffeomorphism.
Obviously $f(\mathbb{R}) = g(\mathbb{R}) = \mathbb{R}$ are homeomorphic as topological spaces. Why does this not count as an "exotic line"?, which is precisely when we have homeomorphic objects but not diffeomorphic ones, the classification theorem says there can be no such thing as an exotic $\mathbb{R}$, since the line is $1$ dimensional and weird things can only happen in starting at dimension $4$.
Perhaps I am struggling to grasp the difference since I have not actually seen what an exotic differential structure looks like, I suppose they require all sorts of advanced constructions and can't be expressed with a nice example, so what makes being exotic so much more bad than just being incompatible?
There is nothing exotic here. Yes, the two atlases give us two distinct smooth structures, i.e. two distinct smooth manifolds with underlying topological space $\mathbb R$. Let us call them $\mathbb R_{id}$ and $\mathbb R_g$. But these smooth manifolds are diffeomorphic: In fact, $g : \mathbb R_g \to \mathbb R_{id}$ is a diffeomorphism since $$\mathbb R \stackrel{g^{-1}}{\to} \mathbb R \stackrel{g}{\to} \mathbb R \stackrel{id}{\to} \mathbb R \phantom{x} , \phantom{x} \mathbb R \stackrel{id^{-1}}{\to} \mathbb R \stackrel{g^{-1}}{\to} \mathbb R \stackrel{g}{\to} \mathbb R$$ are the identity maps which are smooth.
See $C^{\infty } $ diffeomorphism between 2 nonequivalent manifolds .