Unfortunately You can't just use $D/\Gamma$ because then the boundary $S^1$ will be very badly behaved (e.g., nonexistence of nontangential limit by choosing the appropriate point on each fundamental $4g$-gon).
The usual introductory way is to just use Hilbert space theory with minimal amount of Sobolev spaces thrown in.
$(I+\Delta)^{-1}\colon L^2(M)\to H^2(M)$ is continuous, so $(I+\Delta)^{-1}\colon L^2(M)\to H^2(M)\hookrightarrow L^2(M)$ is compact (Rellich-Kondrakov) self-adjoint (easy) and positive (also easy), hence you have an orthonormal basis of eigenfunctions $h_i$, and $\Delta h_i=\lambda_i h_i$ with $0=\lambda_0\leq\lambda_1\leq\dots\leq\lambda_i\uparrow +\infty$.
Expressing $f=\sum a_ih_i$, you get a solution $\varphi=\sum(a_i/\lambda_i)h_i$.
Note that $\lambda_i=0$ iff $h_i$ is constant (by Green's first identity), so a necessarily and sufficient condition is $\int_M f=0$, and if you want uniqueness you can impose $\int_M\varphi=0$.
Since $\Delta^n\varphi=\Delta^{n-1}f\in C^\infty(M)\subset L^2(M)$ for every $n$, we have $\varphi\in\bigcap_n H^{2n}(M)=C^\infty(M)$.
One cannot trust Wikipedia too much on these matters: Anybody can edit a Wikipedia article. The Wikipedia article is sloppy about distinguishing compact and noncompact surfaces. The Uniformization Theorem has the following equivalent forms:
A. Every simply-connected Riemann surface is biholomorphic to $S^2$ or to ${\mathbb C}$ or to the unit disk $\Delta$ in ${\mathbb C}$.
B. Every connected Riemann surface $X$ is biholomorphic either to $S^2= {\mathbb C}P^1$ (with its standard complex structure), or to the quotient of $U={\mathbb C}$ or of $U=\Delta$ by a group $\Gamma$ of linear-fractional transformations of $U$ acting on $U$ freely and properly discontinuously.
C. Every connected Riemannian surface $(S,g)$ admits a positive smooth function $\lambda$ such that $(S, \lambda g)$ is a complete Riemannian manifold of constant curvature. (Note that the surface $S$ is not required to be oriented.)
Remark. i. In all three formulations, the surface $S$ is not required to be compact. The compactness assumption made by the Wikipedia article is totally unnecessary. Compactness is used in some of the proofs but not in other proofs.
ii. There is no way to prove the UT for general noncompact surfaces from the UT for compact surfaces.
Equivalence of the three statements (A, B and C) is not hard to establish. The key facts are the existence of a conformal Riemannian metric on every Riemann surface and that every group $\Gamma$ as in Part B, acts on $U$ isometrically with respect to the Euclidean or hyperbolic metric respectively.
Regarding the Ricci flow, what is true is that there is a proof of the Uniformization Theorem (UT) for compact (closed) Riemann surfaces via the Ricci Flow (RF). This result by itself does not imply the full UT.
In order to prove the UT for noncompact surfaces using RF one would have to work much harder than in the compact case and I am unaware of such a proof in the literature. Even short-term existence of the flow becomes a problem. See for instance
Xiaorui Zhu, Ricci Flow on Open Surface, J. Math. Sci. Univ. Tokyo 20 (2013), 435–444.
for some partial results on proving UT via RF on open surfaces.
Of course, if your (say, simply connected) Riemann surface is compact, then RF indeed does the job: First equip your surface $X$ with an arbitrary conformal Riemannian metric $g_0$ (i.e. a metric which in the local holomorphic coordinates of $X$ has the form $\rho_k(z)|dz|^2$). For a proof of existence of such a metric see my answer here.
Then apply the normalized RF to $g_0$. This, in finite time, converges to a constant curvature metric $g_T$ (the curvature has to be positive). For surfaces, RF preserves the conformal class of the metric. Hence, $g_T$ is still a conformal Riemannian metric on $X$. After rescaling, $g_T$ has curvature $1$. Now, use the theorem (due to Killing and Hopf in all dimensions) that all compact simply-connected surfaces of curvature $1$ are isometric to each other. Hence, $(X,g_T)$ is isometric to the standard unit sphere $S^2$. The isometry $f: X\to S^2$ has to be conformal in the sense of Riemannian geometry, hence (after changing orientation if needed) is conformal in the sense of complex analysis. qed
Best Answer
The italicized statement is quite trivial once you know about the unit disk model of the hyperbolic plane; in contrast, the uniformization theorem is highly nontrivial. As for your question in the last paragraph: Each oriented Riemannian 2-manifold $M$ has a canonical complex structure making it a Riemann surface $X=X(M)$; under this correspondence $M\mapsto X(M)$, every orientation preserving isometry is biholomorphic. In other words, you have a functor from the category of oriented Riemannian surfaces (where morphisms are orientation preserving local isometries) to the category of Riemann surfaces (where morphisms are holomorphic maps).
Every simply connected surface is orientable, but this is not the same as to say that every simply connected surface is oriented. (The concepts oriented and orientable are frequently conflated but that's an abuse of language.)