In   many variational problems one is given an action $S\mapsto S[f]$, where  $S[f]$ as some integral description

$$ S[f]=\int_\Omega L\bigl(\;x,f(x),D f(x),\dotsc, D^k f(x)\;\bigr) dx $$

where $\Omega$ is a region in some Euclidean space $\mathbb{R}^n$, $x\in\Omega$,  $f$ is a  $k$-times differentiable  function $f: \Omega\to\mathbb{R}^m$,  and the Lagrangial $L$ is a function of the appropriate number of variables.  For example  in   classical mechanics $\Omega$  is an interval, $f:\Omega\to \mathbb{R}^m$ describes a path in  $\mathbb{R}^m$ and the Lagrangian  has the form $L(y,v)=K(v)-U(y): \mathbb{R}^m\times\mathbb{R}^m\to \mathbb{R}$, i.e., $\newcommand{\bR}{\mathbb{R}}$

$$L(y,v)=\frac{1}{2}|v|^2-U(y), (y,v)\in\bR^m\times\bR^m $$

so that


$$ L(f,\dot{f})= \frac{1}{2}|\dot{f}|^2-U\bigl(f\bigr), $$ 

where the dot indicates  the derivative with respect to the time parameter $t$ on $\Omega\subset\mathbb{R}$.

The functional (or variational) derivative of $S$ calculated at  $f_0:\Omega\to\mathbb{R}^m$ is a a gadget $\delta S[f_0]$ that  feeds on an infinitesimal deformation $\delta f$ of $f_0$ and returns a scalar


$$ \langle \delta S[f_0], \delta f\rangle =\lim_{h\to 0}\frac{1}{h} \bigl(S[f_0+h\delta f]-S[f_0]\;\bigr). \tag{1}$$

The deformation $\delta f$ is also a function $\Omega\to\mathbb{R}^m$. It is often desirable  to identify  $\delta S[f_0]$ with a function  $g:\Omega\to\mathbb{R}^m$   which, if it exists, is uniquely determined by the equality. 

Making the same formal replacement $\delta f\to\delta(x-x_0)$

$$ \langle \delta S[f_0], \delta f\rangle=\langle g(x), \delta f(x)\rangle =\int_\Omega \bigl( g(x), \delta f(x) \bigr) dx,\tag{2}  $$

where $(-,-)$ denotes the natural inner product on $\mathbb{R}^m$.  The value of $g$ at $x_0$ can be obtained from the  equality

$$ g(x_0)= \langle g, \delta(x-x_0)\rangle. \tag{3} $$

This means that the value of $g$ at $x_0$ is obtained by formally replacing $\delta f$ with $\delta(y-x_0)$ in (2).

Making the same formal replacement $\delta f(x)\to\delta(x-x_0)$ one obtains  the physicists' functional derivative in  your question.

How does one identify $\delta S[f_0]$ with a function?   In the example from classical mechanics  one has


$$ S[f_0](t)=\frac{d}{dt}\frac{\partial L}{\partial v}(f_0(t), \dot{f_0}(t))-\frac{\partial L}{\partial x}(f_0,\dot{f}_0). $$

A good place to look for more details is the book "*Calculus of Variations*" by Gelfand and Fomin,Dover 2000.