Using the definition of the Riemann curvature tensor
$$R_{XY}Z=\nabla_{[X,Y]}Z-[\nabla_X,\nabla_Y]Z,$$
we can compute that
$$R^a_{\;bcd}=\partial_d\Gamma^a_{cb}-\partial_c\Gamma^a_{db}+\Gamma^a_{de}\Gamma^e_{bc}-\Gamma^a_{ce}\Gamma^e_{bd},$$
(note that my convention is $R_{\partial_c\partial_d}\partial_b=R^a_{\;bcd}\partial_a$). Now using our definitions of the Ricci curvature
$$R_{ab}=R^c_{\;acb}$$
and of the scalar curvature
$$R=R^a_{\;a},$$
you would just substitute and calculate.
I'm not sure I understand your second question, since anything expressed in terms of Christoffel symbols is also an expression in coordinates (recall that $\nabla_{\partial_a}\partial_b=\Gamma^c_{ab}\partial_c$).
$\newcommand{\R}{\mathbb{R}}$
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
Third Edit (the charm??): I don't know what Chow and Knopf had in mind when they wrote the equation $|\text{Rm}(t)| = CR$, because it's just wrong. But there's a very simple argument to show that the Ricci flow on a positive-scalar-curvature Einstein manifold has a Type I singularity.
Since the evolution of the meetric has the form $g(t) = a(t) g_0$ for some positive function $a(t)$, the inverse metric, $(0,4)$-Riemann curvature tensor, and scalar curvature transform as follows: $$\begin{align} \text{Rm}(x,t) &= a(t)\,\text{Rm(x,0)},\\ R(x,t) &= a(t)^{-1} R(x,0),\\ g(x,t)^{-1} & = a(t)^{-1} g_0(x)^{-1}. \end{align} $$ (See Theorem 7.30 in my Introduction to Riemannian Manifolds.) Thus the squared norm of $\text{Rm}(x,t)$ satisfies $$ \begin{align} \|\text{Rm}(x,t)\|^2 &= g(x,t)^{-1}*g(x,t)^{-1}*g(x,t)^{-1}*g(x,t)^{-1} * \text{Rm}(x,t)*\text{Rm}(x,t) \\ &=a(t)^{-2}\|\text{Rm(x,0)}\|^2 \end{align} $$ (where the asterisks represent contractions on various indices). Plugging in $a(t) = C(T-t)$ shows that $\|\text{Rm}(x,t)\|(T-t)$ is constant for each $x$. Since a complete Einstein manifold with positive scalar curvature must be compact by Myers's theorem, this quantity is bounded on $M$.
Here's why $|\text{Rm}(t)| = CR$ (or maybe they meant $CR(t)$) is not true no matter how you interpret it. If you interpret the left-hand side as a pointwise norm, it can't be correct because $R(t)$ is constant for each $t$ while $|\text{Rm}(t)|$ is not. If you interpret $|\text{Rm}(t)|$ to mean the global $L^2$ norm (i.e., the square root of the integral of the squared pointwise norm), then it's not true either because when $g(t) = a(t) g_0$ we have $R(t) = a(t)^{-1}R(0)$ while $$ \left(\int_M |\text{Rm}(t)|^2dV_{g(t)}\right)^{1/2} = a(t)^{-1+n/4}\left(\int_M |\text{Rm(0)}|^2dV_{g_0}\right)^{1/2}. $$ (See Theorem 7.30 in IRM.)