$\newcommand{\R}{\mathbb{R}}$
The basic heat equation on a domain in $\R^n$ is
$$ \partial_tu = \sum_{i=1}^n (\partial_i)^2u $$
The inhomogeneous version is
$$ \partial_tu = \sum_{i=1}^n (\partial_i)^2u + f, $$
where $f$ is a function of $x \in \R^n$ only.
This all can be generalized to the linear PDE
$$ \partial_tu = a^{ij}(x,t) \partial^2_{ij}u + b^k(x,t)\partial_ku + c u+ f(x), $$
where the matrix $A = [ a_{ij}]$ is positive definite.
This is the most general version of what is often called a linear heat equation on a domain in $\R^n$.
A quasilinear heat equation is simply one where the functions $a^{ij}$, $b^k$, $c$, and $f$ are functions of not only $x$, $t$, but also $u$.
On a Riemannian manifold, one possible linear heat equation is of the form
$$
\partial_tu = \Delta_g u + b^k\partial_k u + cu + f,
$$
where $\Delta_g u = g^{ij}\nabla^2_{ij}u$.
All of the above assumes $u$ is a scalar function on its domain.
Where things get more complicated is when $u$ is either a map $u: M \rightarrow N$, as it is for the harmonic map heat flow or the Riemannian metric itself, as it is for the Ricci flow. If you write out the formulas for these PDEs in local coordinates, you will see that the coefficients of the PDE depend on the unknown map or metric and therefore the PDE is nonlinear. The fact that the harmonic map heat flow can be written simply as
$$
\partial_t u = \Delta u
$$
is misleading, because here $\Delta u$ is a nonlinear function of its first and second partial derivatives.
As for why the Ricci flow is a nonlinear heat equation is longer story, but after you use the so-called DeTurck trick, the Ricci flow looks like
$$
\partial_t g_{ij} = g^{pq}\partial^2_{pq}g_{ij} + \text{ lower order terms}
$$
which is a nonlinear heat equation for the metric $g$.
There are similar stories for other geometric heat flows, such as the mean curvature flow.
Best Answer
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