Variation of a field action

According to Ortin, Gravity & Strings (Chapter 2), we define

$$dfrac{delta S}{delta phi} equiv dfrac{partial mathcal{L}}{partial phi} - partial_mu bigg(dfrac{partial mathcal{L}}{partial(partial_muphi)}bigg)$$

Now, as you can easily see,

$$delta S = displaystyleint d^4x bigg(dfrac{partial mathcal{L}}{partial phi} delta phi - dfrac{partial mathcal{L}}{partial(partial_muphi)} delta (partial_mu phi)bigg) $$

$$= displaystyleint d^4x bigg(dfrac{partial mathcal{L}}{partial phi} - partial_mu bigg(dfrac{partial mathcal{L}}{partial(partial_muphi)}bigg)bigg) delta phi $$

$$=displaystyleint d^4x dfrac{delta S}{delta phi} delta phi$$

Edit From this answer, I realize that we don't really need to define an explicit $dfrac{delta S}{delta phi}$. Rather, we can derive it from the definition of $S$ as it should be the case. Based on the answer linked above, I derive here $dfrac{delta S}{delta phi}$ in the simplest manner apparent to me. The key is to realize that there is an implicit label associated with $phi$ which decides what $phi$ we are talking about - the label is the coordinates $x$.

$$S=displaystyleint d^4x mathcal{L}(phi(x), partial_mu phi(x)) $$ $$delta S = displaystyleint d^4x bigg(dfrac{partial mathcal{L}}{partial phi (x)} delta phi(x) - dfrac{partial mathcal{L}}{partial(partial_muphi(x))} delta (partial_mu phi(x))bigg) $$

$$= displaystyleint d^4x bigg(dfrac{partial mathcal{L}}{partial phi(x)} - partial_mu bigg(dfrac{partial mathcal{L}}{partial(partial_muphi(x))}bigg)bigg) delta phi(x) $$

Now, since $x$ is already used in the integration and it runs over all of spacetime, we should use a different variable for coordinate label when we want to define $dfrac{delta S}{delta phi}$ which is the variation of action with respect to variation in the field at any point in spacetime. Let's use $y$ to denote coordinates of this point where we vary the field and notice the variation in action.

$$dfrac{delta S}{delta phi} = dfrac{delta S}{delta phi (y)}$$ $$= displaystyleint d^4x bigg(dfrac{partial mathcal{L}}{partial phi(x)} - partial_mu bigg(dfrac{partial mathcal{L}}{partial(partial_muphi(x))}bigg)bigg) dfrac{delta phi(x)}{delta phi(y)} $$ $$= displaystyleint d^4x bigg(dfrac{partial mathcal{L}}{partial phi(x)} - partial_mu bigg(dfrac{partial mathcal{L}}{partial(partial_muphi(x))}bigg)bigg) delta (x -y) $$ $$=dfrac{partial mathcal{L}}{partial phi(y)} - partial_mu bigg(dfrac{partial mathcal{L}}{partial(partial_muphi(y))}bigg)$$

So, $$dfrac{delta S}{delta phi(y)} = dfrac{partial mathcal{L}}{partial phi} - partial_mu bigg(dfrac{partial mathcal{L}}{partial(partial_muphi)}bigg)$$ And, thus, $$delta S = displaystyleint d^4x bigg(dfrac{partial mathcal{L}}{partial phi(x)} - partial_mu bigg(dfrac{partial mathcal{L}}{partial(partial_muphi(x))}bigg)bigg) delta phi(x) $$ $$=displaystyleint d^4x dfrac{delta S}{delta phi(x)} delta phi(x)$$

Or, concisely, ${delta S} = displaystyle int d^4x dfrac{delta S}{delta phi} delta phi$.

I guess the confusion is caused by (implicitly) different variables and their associated derivatives. I'll stick to the action of a single classical scalar real field in the following, and try a heuristic explanation.

• In a global view, the action is a function of the field, $S!left(phiright)$. Here, $phi$ denotes the 'complete' field, i.e. the function $$phi: mathbb{R}^4to mathbb{R}$$ (or potentially a subset of $mathbb{R}^n$, and $mathbb{C}$ on the right). Hence, $phi$ is a single variable, but taken from an infinite-dimensional space, and we can think about the derivative $$frac{delta S(phi)}{delta phi},,$$ measuring how much $S$ varies as $phi$ is changed. This is again a number for each field $phi$, i.e. a function on the space of fields. We're using $delta$'s here to point out that it is not a simple partial derivative (this is historical, similar to the use of square brackets and the name 'functional'). Note that at this point, it doesn't make much sense to put coordinates on the $phi$, nor to separately differentiate with respect to derivatives of $phi$ - if the function $phi$ is given, its derivatives are fixed.
• From a local/Lagrangean point of view, we also assume that the action is given by a single spacetime integral as $$S(phi)=int text{d}^4x,mathscr{L}!left(phi(x),partial_muphi(x)right),.$$ Here $mathscr{L}$ is a function of $phi$ and its derivatives, all evaluated at the same point $x$ (that's locality, basically). In other words, $mathscr{L}$ can be thought of as a function of five real numbers (the field and four derivatives). Now the variation of the action can be expressed more explicitly as $$frac{delta S(phi)}{delta phi}=int text{d}^4x, left(frac{partialmathscr{L}!left(phi(x),partial_muphi(x)right)}{partial(phi(x))} - partial_mufrac{partialmathscr{L}!left(phi(x),partial_muphi(x)right)}{partial(partial_mu phi(x))}right),.$$ Here, the derivative in the integral are simple partial derivatives of the function $mathscr{L}$ with respect to its five arguments.
• Finally, we can have a mixed viewpoint, basically a variation of the first one: We consider the field $phi$ to be a collection of uncountably many field values $phi(x)$, so that $x$ is just a label (similar to how a function of a vector $vec{v}$ can be regarded as a function of $d$ numbers $v_i$ labelled by the index $i$). Now we can even differentiate $S$ with respect to field values at given points. For an action given by the integral of a Lagrangean, we have $$frac{delta S(phi)}{delta phi(x)}=frac{partialmathscr{L}!left(phi(x),partial_muphi(x)right)}{partial(phi(x))} - partial_mufrac{partialmathscr{L}!left(phi(x),partial_muphi(x)right)}{partial(partial_mu phi(x))},,$$ i.e. the derivative wrt $phi(x)$ is an expression localised at $x$. Combining this expression and the previous one, we can write the variation of the action as $$delta S= int text{d}^4x,frac{delta S(phi)}{delta phi(x)} delta phi(x),.$$ When you drop the $x$'s, this looks almost like the first expression - but you still integrate over $x$, so the argument is a function of spacetime.

The different notions of variables, $delta$'s and $partial$'s (and later $mathscr{D}$'s!) can be confusing, but usually it becomes clear from the context what is meant.