differential topology - exotic smooth structure clarification

What do you mean by "smooth mapping" of atlases? I think you maybe using the terminology a bit different from how I do.

A chart or a coordinate chart on a topological manifold $M$ is a pair $(U,psi)$ where $Usubset M$ is open and $psi: U to Vsubset mathbb{R}^n$ is a homeomorphism. Given two charts $(U_1,psi_1)$ and $(U_2,psi_2)$ with non-empty intersection $U_1cap U_2$, we say that the charts are $C^k$ compatible if the transition function $psi_2circpsi_1^{-1} |_{psi_1(U_1cap U_2)}$ is $k$-times continuously differentiable. Then a $C^k$ atlas on $M$ is a collection of charts ${(U_alpha,psi_alpha)}_{alpha in A}$ such that the union $cup_{alpha in A} U_alpha$ covers $M$ and all the charts are $C^k$-compatible.

Two atlases are said to be $C^k$ compatible if all of their corresponding charts are pair-wise compatible, and hence if $mathcal{A},mathcal{B}$ are two compatible atlases, their union also is an atlas. An atlas is said to be maximal if any other compatible atlas must be a subset. It always exists by Zorn's lemma.

A differential structure on the manifold $M$ is a choice of a maximal $C^
infty$ atlas, we write it as $(M,mathcal{A})$. Two differentiable manifolds $(M,mathcal{A})$ and $(N,mathcal{B})$ are said to be diffeomorphic if there exists a homeomorphism $Psi: Mto N$ such that for any $(U_{alpha},psi_alpha) in mathcal{A}$ and $(V_{beta}, phi_{beta})inmathcal{B}$ we have that the function
$$ phi_beta circ Psi circ psi_alpha^{-1} |_{psi_alpha(U_alpha cap Psi^{-1}(V_beta))} $$
is smooth.

Ryan already gave you an answer to your question. But let me elaborate a bit on your question two. Let $mathbb{R}^2$ be your manifold. Define two charts on it
$$ U_1 := mathbb{R}times (-1,infty), U_2 := mathbb{R}times (-infty,1) $$
and define $psi_1(x) = x$ if $xin mathbb{R}times (-1,2)$ and $psi_1(x_1,x_2) = (x_1 + x_2 - 2, x_2)$ if $x_2 geq 2$. Similarly $psi_2$. Clearly $(U_1,psi_1), (U_2, psi_2)$ cover $mathbb{R}^2$ and the transition function on the strip $mathbb{R}times (-1,1)$ is equal to the identity, and hence is smooth, so this gives a smooth atlas. But this smooth atlas is not compatible with the standard Cartesian atlas. What that does not mean that $mathbb{R}^2$ with this atlas is exotic!

Let $(mathbb{R}^k,mathcal{E})$ denote the standard Euclidean space. An exotic smooth structure $(mathbb{R}^k,mathcal{A})$ requires that every homeomorphism from $mathbb{R}^k$ to itself to not extend to a diffeomorphism from $(mathbb{R}^k,mathcal{E})to (mathbb{R}^k,mathcal{A})$. What we have here is just that our stupid choice of homeomorphism is non-smooth. It is simple to change our homeomorphism such that the mapping is smooth relative to the fixed atlases.

So the non-existence of exotic smooth structure just say that for any two fixed smooth atlases, there exists some homeomorphism from $mathbb{R}^k$ to itself such that it extends to a diffeomorphism. If this is what you meant, then the answer to your second question is "yes". But again, I don't understand what you mean by "smooth mapping of atlases".