What is the infinite-dimensional-manifold structure on the space of smooth paths mod thin homotopy?
Okay, you asked for it!
Question: What is the manifold structure on $P^1(M)$?
Answer: There isn't one.
Update: The biggest failing is actually that the obvious model space is not a vector space. The space of paths mod thin homotopy in $\mathbb{R}^n$ does not inherit a well-defined addition from the space of paths in $\mathbb{R}^n$. Full details at the nLab page http://ncatlab.org/nlab/show/smooth+structure+of+the+path+groupoid.
(Update added here, rather than at the end, as it's the most direct answer to the specific question; the rest should be viewed as extra for those interested in more than just whether or not this space is a smooth manifold.)
It is, as you say, a smooth space. This is formal: whatever category of generalised smooth spaces you like, take the quotient of $P(M)$ by thin homotopies. All the proposed categories of generalised smooth spaces admit quotients, so the quotient exists and is a smooth space. Depending on your choice of category, the description of this smooth space may vary. For example, its Frolicher structure and its diffeological structure are very different.
But it is not "locally linear" in any sense. The basic problem is that, as you say, within an equivalence class you have paths wrapping all the way around the manifold. This destroys any hope of local linearity.
As for the proposed local model, you hit the nail on the head when you say:
It's not clear to me how to put a topology on it,
Absolutely! Topologising these spaces can lead to quite strange behaviour. You want a LCTVS structure, else you haven't a hope of even starting, and that can distort the topology from what you expect. For example, if you take piecewise-smooth paths (with no quotient) then the LCTVS topology on that is the $C^0$-topology! Indeed, simply taking so-called "lazy paths" could be fraught with difficulties (I notice that you define "lazy" slightly differently to how I've seen it done before with sitting instances). Is that space a manifold? (I know the answer to this one, but if you don't then you should start with that one as it is a much easier question and will hone your skills a little.)
If you really want a manifold, the solution is to go one step further. Rather than quotienting out by thin homotopies, make your "thing" into a 2-structure and put the thin homotopies in at the 2-level. Keep all paths at the 1-level. Then each level has a manifold structure and by mapping into a 1-structure you effectively quotient out by the 2-structure but never actually have to consider the quotient itself.
To coin a phrase:
Quotients are horrible, it's a shame so many people think otherwise.
Lastly, that's not to say that there is no way of making $P^1(M)$ into a manifold. There may well be. But if there is, it'll be so convoluted and contrived that it won't look anything like the quotient of $P(M)$. A cautionary tale here is the case of all paths in a manifold, $C^\infty(\mathbb{R},M)$. That can be made into a manifold, but it has uncountably many components, for example, so looks absolutely horrid.
Okay, not quite lastly. There's lots of details here that have been glossed over. If you are really interested in working out the smooth space structure of this particular space then I (and I suspect Urs and Konrad) would be very interested in seeing it done and helping out. But MO isn't the place for that. Hop on over to the nLab, create a spin-off of http://ncatlab.org/nlab/show/path+groupoid, and start working.
Further Reading
Constructing smooth manifolds of loop spaces, Proc. London Math. Soc. 99 (2009) pp195–216 (doi:10.1112/plms/pdn058, arXiv:math/0612096). The point of this is to figure out exactly when the "standard method" (alluded to by Tim) works. The distinction between "loop" and "path" is irrelevant.
The Smooth Structure of the Space of Piecewise-Smooth Loops, Glasgow Mathematical Journal, 59 (2017) pp27-59. (arXiv:0803.0611, doi:10.1017/S0017089516000033). Why you should be very, very nervous whenever anyone says "consider piecewise-smooth maps"; and take as a cautionary tale as to the inadvisability of going beyond smooth maps in general.
Work of David Roberts on the nLab. This is where I got the 2-idea that I mentioned above.
Other relevant nLab pages: http://ncatlab.org/nlab/show/generalized+smooth+space, http://ncatlab.org/nlab/show/smooth+loop+space and further.
Of course, the magnificent book by Kriegl and Michor. (I'm going to create a separate MO account for that book; its role will be to post an answer on relevant questions simply saying "Read Me".)
In response to Konrad's comment below, I've started an nlab page to work out the smooth structure of this space. The initial content considers the linear structure of the space of paths in some Euclidean space modded out by thin homotopy. The page is http://ncatlab.org/nlab/show/smooth+structure+of+the+path+groupoid.
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 $\dot{\gamma}(t)$ and $\xi(t)$ are linearly dependent in $T_{\gamma(t)}M$ 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, $g$ 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.