How to define $\delta{g_{\mu\nu}}$? In general relativity, when deriving the field equation using the variational principle we use $\hat{g}_{\mu\nu}=g_{\mu\nu}+\delta{g_{\mu\nu}}$.
Does $\delta{g_{\mu\nu}}$ mean the measurement of how $g_{\mu\nu}$ changes when we change the form of the equations in the components of ${g_{\mu\nu}}$ by changing the coordinates or doing some other thing  that changes their form, Or does it measure how $g_{\mu\nu}$ changes when we translate $r$ by $\triangle{r}$?
 A: Classical Lagrangian field theory deals with fields $\phi: M \to N$, where $M$ is spacetime and $N$ is the target-space of the fields. We shall for convenience call $M$ and $N$ the horizontal and the vertical space, respectively. The metric $g$ can be viewed as a classical field of this kind.
OP is asking about finding the Euler-Lagrange equations. In that case, the variations are vertical.
There are other applications where variations are not necessarily vertical, e.g. Noether's theorem, cf. this Phys.SE post. 
A: Usually, the variation of a filed $\delta\phi$ is defined to be 
\begin{equation}
\delta\phi\left(x\right)=\phi^{'}\left(x\right)-\phi\left(x\right)
\end{equation}
where the new field $\phi^{'}$ and old field $\phi$ are evaluated
at the same point, if we take a active transformation point of view.
So I think $\delta g_{\mu\nu}$ should be 
\begin{equation}
\delta g_{\mu\nu}\left(x\right)=\hat{g}_{\mu\nu}\left(x\right)-g_{\mu\nu}\left(x\right)
\end{equation}
where they are evaluated at the same point.
So if you make a translation, $x_{\mu}\rightarrow x_{\mu}+\epsilon_{\mu}$,
then $\hat{g}_{\mu\nu}\left(x+\epsilon\right)=g_{\mu\nu}\left(x\right)$.
But 
\begin{equation}
\hat{g}_{\mu\nu}\left(x+\epsilon\right)=\hat{g}_{\mu\nu}\left(x\right)+\epsilon^{\alpha}\partial_{\alpha}g_{\mu\nu}\left(x\right)
\end{equation}
So by our definition, we have
\begin{align*}
\delta g_{\mu\nu}\left(x\right) & =\hat{g}_{\mu\nu}\left(x\right)-g_{\mu\nu}\left(x\right)\\
 & =\hat{g}_{\mu\nu}\left(x+\epsilon\right)-\epsilon^{\alpha}\partial_{\alpha}g_{\mu\nu}\left(x\right)-g_{\mu\nu}\left(x\right)\\
 & =-\epsilon^{\alpha}\partial_{\alpha}g_{\mu\nu}\left(x\right)
\end{align*}
