A way to look at it:
s, f, g ##\in## M the function space; ##V:M\to \mathbb R## a functional.
- ##< f, g> = \int f(x)g(x)\,dx## is a bilinear, symetric and "almost" non-degenerate form. It associate a fuction with a form: $$f\in M \mapsto <f,\cdot> \,or\,<\cdot,f>\in M^*$$
- The Dirac delta function ##\delta (x) = \delta_0## or ##\delta(x-y) = \delta_y## ##\in## M, y fixed, is the function that correspond to the form $$<\delta_y,\cdot>\in M^*\quad s.t. \quad <\delta_y,f> = f(y)\in \mathbb R$$
- The functional derivative of a functional at the point(=function) ##g\in M## is the function ##\frac{\delta V}{\delta g}## or ##\frac{\delta V}{\delta g(x)}## (as the notations f or f(x)) such that: $$<\frac{\delta V}{\delta g},\cdot>\in M^* = \mathbf D_g V(\cdot)$$ ##D_g V(\cdot)## the derivative of V at g. It's a linear map sending a vector h at g (h=function again) to the directional derivative of V at g in the direction h:$$D_g V(\cdot): h\mapsto D_g V(h) = \lim_{\epsilon \rightarrow 0}{\frac{V(g+\epsilon h)-V(g)}{\epsilon}}\in \mathbb R$$
Then ##<\delta_y,\frac{\delta V}{\delta g}> = \frac{\delta V}{\delta g(y)}## the value of the function ##\frac{\delta V}{\delta g(x)}## at x=y.