If you want to prove something is a smooth manifold, a good way to begin is to decide what its tangent spaces ought to be. So let $\gamma_s$ be a (smooth) homotopy of lazy paths, say for $s$ in $(-\epsilon,\epsilon)$. Its derivative at $s=0$ is a vector field $\xi$ along $\gamma:=\gamma_0$. This is a section of $\gamma^\ast TM$, not necessarily a "lazy" one. The vector field is to count for nothing if $\gamma_s$ is a lazy thin homotopy. So we should take the quotient of $C^\infty(\gamma^* TM)$ by the subspace $L$ of those $\xi$ which have vanishing (higher) derivatives at $0$ and $1$ and such that $\dot{\gamma_0}(t)$ \dot{\gamma}(t)$and$\xi(t)$are linearly dependent in$T_{\gamma_0(t)}M$T_{\gamma(t)}M$ for all $t$.
A standard method to produce smooth charts (on path spaces in particular) is to exponentiate tangent vectors. This requires an auxiliary choice, say of a metric $g$ on $M$, so the manifold structure won't be absolutely canonical; but it may well be canonical up to diffeomorphism (strategy: define a smooth structure on the family of manifolds parametrized by the contractible manifold $Met(M)$, and show it's a smooth fibre bundle).
Well, $\gamma$ g$induces an$L^2$-metric on$C^\infty(\gamma^{\ast}TM)$, so we could take the orthogonal complement$L^{\perp}$(isn't that the vector fields pointwise-orthogonal to$\dot{\gamma}$, vanishing where$\dot{\gamma}$does?) and view that as our tangent space. That makes it a little clearer that it's a Frechet space (consider the$C^k$norms on$L^\perp$...). Let$L^{\perp}_\epsilon$be the vector fields in$L^\perp$which, pointwise, have length$<\epsilon$. Assume$\epsilon$is smaller than the injectivity radius of$g$along$\gamma$. Then one has$Exp_g \colon L^\perp\to \mathcal{P}^1 M$(since it defines a diffeo from$(T_{\gamma(0)}M)_{<\epsilon}$onto its image,$\exp_g$preserves laziness). This map is injective, and it's a reasonable candidate for a coordinate chart. Declare such charts to be our atlas, defining, as a by-product, a topology - the coarsest that makes the charts continuous. Now you have several things to check. (I haven't - maybe it doesn't work...) One of those is that the topology is Hausdorff, so you might even want to make this into a metric space, perhaps via a Riemannian metric. 1 If you want to prove something is a smooth manifold, a good way to begin is to decide what its tangent spaces ought to be. So let$\gamma_s$be a (smooth) homotopy of lazy paths, say for$s$in$(-\epsilon,\epsilon)$. Its derivative at$s=0$is a vector field$\xi$along$\gamma:=\gamma_0$. This is a section of$\gamma^\ast TM$, not necessarily a "lazy" one. The vector field is to count for nothing if$\gamma_s$is a lazy thin homotopy. So we should take the quotient of$C^\infty(\gamma^* TM)$by the subspace$L$of those$\xi$which have vanishing (higher) derivatives at$0$and$1$and such that$\dot{\gamma_0}(t)$and$\xi(t)$are linearly dependent in$T_{\gamma_0(t)}M$for all$t$. A standard method to produce smooth charts (on path spaces in particular) is to exponentiate tangent vectors. This requires an auxiliary choice, say of a metric$g$on$M$, so the manifold structure won't be absolutely canonical; but it may well be canonical up to diffeomorphism (strategy: define a smooth structure on the family of manifolds parametrized by the contractible manifold$Met(M)$, and show it's a smooth fibre bundle). Well,$\gamma$induces an$L^2$-metric on$C^\infty(\gamma^{\ast}TM)$, so we could take the orthogonal complement$L^{\perp}$(isn't that the vector fields pointwise-orthogonal to$\dot{\gamma}$, vanishing where$\dot{\gamma}$does?) and view that as our tangent space. That makes it a little clearer that it's a Frechet space (consider the$C^k$norms on$L^\perp$...). Let$L^{\perp}_\epsilon$be the vector fields in$L^\perp$which, pointwise, have length$<\epsilon$. Assume$\epsilon$is smaller than the injectivity radius of$g$along$\gamma$. Then one has$Exp_g \colon L^\perp\to \mathcal{P}^1 M$(since it defines a diffeo from$(T_{\gamma(0)}M)_{<\epsilon}$onto its image,$\exp_g\$ preserves laziness). This map is injective, and it's a reasonable candidate for a coordinate chart. Declare such charts to be our atlas, defining, as a by-product, a topology - the coarsest that makes the charts continuous.