Manifold of smooth mappings

Remarks

Each $\varphi_f$ is a bijection onto an open subset of $\Gamma(f^* TN)$, with inverse $s \mapsto \exp \circ s$, and the chart changes are smooth. The resulting manifold structure does not depend on the choice of the metric, its topology is the $C^\infty$ topology, and the tangent space at $f$ is $\Gamma(f^* TN)$.

Example: for $M = N = S^1 = \R / 2\pi\Z$ with the flat metric, take $V$ to be the vectors of length less than $\pi$. For the identity map $f$, the chart sends $g$ near $f$ to the function $s$ with $g(x) = x + s(x)$ and $\abs{s(x)} < \pi$. So near the identity, the diffeomorphisms of the circle correspond to small smooth functions on the circle.

Non-example: in the same setting, let $f$ be the constant map at a point $y$, and $g$ the constant map at the antipodal point $y + \pi$. Then $(f(x), g(x)) = (y, y + \pi)$ is a pair at distance $\pi$, so it lies outside $W$, so $g$ is not in the chart domain $U_f$; it lies in a different chart.

When $\dim M \ge 1$ and $\dim N \ge 1$, the model spaces are Fréchet spaces that are not normable, so this atlas is not an atlas of a Banach manifold.