For a function $f$ of a real variable $t$ a small increment at $p$ is just a real number $\epsilon$ with $|\epsilon|\ll1$. In order to find conditions that permit a local extremum of $f$ at $p$ we therefore have to study
$$f(p+\epsilon)-f(p)=f'(p)\epsilon+o(\epsilon)\qquad(\epsilon\to0)$$
and find that $f'(p)=0$ is a necessary condition (This is Calculus 101). Note that, up to sign, there is just one possible direction for a small increment.
Now in an $n$-dimensional, or even $\infty$-dimensional environment a "tiny increment" at $p$ is best described as $\epsilon u$, whereby $|\epsilon|\ll1$, and $u$ is an arbitrary fixed vector making sense in said environment. We then look at the increments $f(p+\epsilon u)-f(p)$ in direction $u$. If $p\in{\mathbb R}^n$ then, according to multivariable calculus,
$$f(p+\epsilon u)-f(p)=\epsilon\> \nabla f(p)\cdot u+o(\epsilon)\qquad(\epsilon\to0)\ .$$
When $|\epsilon|$ is small this will assume values of both signs for certain $u$ unless $\nabla f(p)=0$. It follows that $\nabla f(p)=0$ is a necessary condition for a local extremum of $f$ at $p$.
Similar arguments apply if $f(x)$ denotes a certain integral involving the function $x:\>[a,b]\to{\mathbb R}$. In this case both $p$ and $u$ are functions defined on $[a,b]$. The study of such a situation then leads to the Euler-Lagrange condition: In the end all "directional derivatives" at $p$ have to be zero whatever the "increment direction" $u$.