Variation in Einstein-Hilbert action In this page there are calculations of variation of Einstein-Hilbert action.
I see variations of terms like this: 

$\delta {R^{\rho }}_{{\sigma \mu \nu }}$

where the term is not a functional, and 

$\frac  {\delta {\mathcal  {L}}_{{\mathrm  {M}}}}{\delta g^{{\mu \nu }}}$

where we have a functional derivative of a term that is not also a functional.
What is the exact meaning of those expressions?
 A: The symbol "$\delta$" in calculus of variation, especially in physics, denotes the change of an output variable due to the change of input variable(s). In general, let $g:Y_1\times Y_2\times\cdots\times Y_n\to Z$ be a mapping, then
$$
\left(\delta g\right)(y_1,y_2,\cdots,y_n):=g(y_1+\delta y_1,y_2+\delta y_2,\cdots,y_n+\delta y_n)-g(y_1,y_2,\cdots,y_n).
$$
Here are some examples.
(1) Consider $f:\mathbb{R}\to\mathbb{R}$, which is a function. Then
$$
\left(\delta f\right)(x)=f(x+\delta x)-f(x)=f'(x)\delta x+o(\delta x),
$$
which is exactly the differentiation of this function, i.e.,
$$
\left({\rm d}f\right)(x)=f(x+{\rm d}x)-f(x)=f'(x){\rm d}x+o({\rm d}x).
$$
It is conventional to keep only the first-order variation and drop the higher-order terms. Thus
$$
\left(\delta f\right)(x)=f'(x)\delta x.
$$
(2) Consider $J:\mathscr{F}\to\mathbb{R}$ with $\mathscr{F}$ being some function space, which is a functional. Then
$$
\left(\delta J\right)(f)=J(f+\delta f)-J(f).
$$
(3) Consider $g=\det\left(g_{\mu\nu}\right)$. Recall the definition of the determinant, and $g$ can be regarded as a polynomial function of all $g_{\mu\nu}$'s. For example,
$$
\left(g_{\mu\nu}\right)=\left(
\begin{array}{cc}
a&b\\
c&d
\end{array}
\right)\Longrightarrow\det g=ad-bc,
$$
which is a second-order multivariate polynomial. In this sense, $g$ is a function of all of its entries, and $\delta g$ is no more than the differentiation of $g$ with respect to those entries, i.e.,
$$
{\rm d}g=g^{\nu\mu}g{\rm d}g_{\mu\nu}\iff\delta g=g^{\nu\mu}g\delta g_{\mu\nu}.
$$
(4) Consider $\mathcal{L}[x(t)]=\left(m/2\right)\dot{x}^2(t)-\Phi_g(x(t))$ with $\Phi_g$ being, say, gravitational potential, which is the integrant of the Newtonian action. In this case, "$\delta$" is in the sense of a shorthand of the respective functional derivative. That is, define
$$
J[x(t)]=\int_{t_1}^{t_2}\mathcal{L}[x(t)]{\rm d}t,
$$
and we have, using integration by parts,
$$
\left(\delta J\right)[x(t)]=\int_{t_1}^{t_2}\left(-m\ddot{x}(t)-\left(\nabla\Phi_g\right)(x(t))\right)\delta x(t){\rm d}t.
$$
With this in mind, denote
$$
\delta\mathcal{L}[x]=\left(-m\ddot{x}-\left(\nabla\Phi_g\right)(x)\right)\delta x.
$$
Remark. In this last example, $\mathcal{L}[\cdot]$ and $\mathcal{L}(\cdot)$ do not mean the same. $\mathcal{L}[\cdot]$ is a mapping defined as above, while $\mathcal{L}(\cdot)$ is simply regarded as a function, e.g., let $\mathcal{L}(p,q)=\left(m/2\right)p^2-\Phi_g(q)$, and $\mathcal{L}[x(t)]=\mathcal{L}(p,q)|_{p=\dot{x}(t),q=x(t)}$.
