Broad definition of functional
Let's write the a function $F(\dot{x}(t), x(t), t)$, whose first two arguments are a function $x(t)$ and it's time derivative $\dot{x}(t)$ with respect to $t$, the independent variable, that could be an argument (the third one, here) of the function $L$.
Being $F$ a function of functions (as the first two arguments), we could call it a functional.
As suggested by @hft, in Physics and in other fields, we're usually interested in functionals defined with a definite integral
\begin{equation}
I[x(t)] = \int_{t \in \Omega} F(\dot{x}(t), x(t), t) dt \ .
\end{equation}
Why functionals and variations?
Now, many problem in Math and Physics can be formulated as a principle of stationarity of a functional with respect to the function argument $x(t)$. How to find that function? Approximately like we do for derivatives:
- introducing a (small) "increment"/variation function $\varepsilon \eta(t)$, satisfying all the constraints of the problem (if any); here the "small" parameter $\varepsilon$ is introduced to make it goes to zero later
- evaluating the difference between the function $L$ with the "variated function argument" $x(t)+\varepsilon \eta(t)$ and the "original functional argument" $x(t)$;
- taking this difference as a function of $\varepsilon$ and performing a series expansion around $\varepsilon = 0$,
\begin{equation}\begin{aligned}
\delta F(\dot{x}(t), x(t), t; \varepsilon) & = F(\dot{x}(t)+\varepsilon \dot{\eta}(t), x(t)+\varepsilon\eta(t),t) -
F(\dot{x}(t), x(t),t) = \\
& = \varepsilon \, \dot{\eta}(t) \, \dfrac{\partial F}{\partial \dot{x}}(\dot{x}(t), x(t),t) + \varepsilon \, \eta(t) \, \dfrac{\partial F}{\partial \dot{x}}(\dot{x}(t), x(t),t) + o(\varepsilon^2) \ \ ,
\end{aligned}\end{equation}
- perform the incremental variation for $\varepsilon \rightarrow 0$
\begin{equation}
\lim_{\varepsilon \rightarrow 0} \dfrac{ \delta F(\dot{x}(t), x(t), t; \varepsilon)}{ \varepsilon} = \dot{\eta}(t) \, \dfrac{\partial F}{\partial \dot{x}}(\dot{x}(t), x(t),t) + \eta(t) \, \dfrac{\partial F}{\partial \dot{x}}(\dot{x}(t), x(t),t) \ .
\end{equation}
Now, and only now (when you understood a bit what we've done so far), I'd suggest to call $\eta(t) = \delta x(t)$ to remind you it's the function we used to vary the function $x(t)$, and $\delta F = \lim_{\varepsilon \rightarrow 0} \frac{ \delta F(\dot{x}(t), x(t), t; \varepsilon)}{ \varepsilon}$, for brevity (and remind you it's the result of a variation). With this notation, we can now re-write
\begin{equation}
\delta F(\dot{x}(t), x(t), t) = \delta \dot{x}(t) \dfrac{\partial F}{\partial \dot{x}}(\dot{x}(t), x(t), t) + \delta x(t) \dfrac{\partial F}{\partial x}(\dot{x}(t), x(t), t) \ .
\end{equation}
Note: variations and differentials.
This last paragraph aims at answering the last question about the equivalence of differentials and variations, and to highlight the difference between them in the absence of the term $\frac{\partial F}{\partial t}$ (and thus completing the comment under @Cleonis answer).
If we take $F(\dot{x}(t), x(t), t)$ as a function of time $\tilde{F}(t) := F(\dot{x}(t), x(t), t)$, its differential reads
\begin{equation}
dF = d \dot{x}(t) \dfrac{\partial F}{\partial \dot{x}}(\dot{x}(t), x(t), t) + d x(t) \dfrac{\partial F}{\partial x}(\dot{x}(t), x(t), t) + \dfrac{\partial F}{\partial t} (\dot{x}(t), x(t), t) \ ,
\end{equation}
and thus, beside the different notation, it differs from the variation of $F$ for the presence of the term $\frac{\partial F}{\partial t} (\dot{x}(t), x(t), t)$.
The reason why this term doesn't appear in the variation of the functional come from the process used to evaluate variation, and in detail from the very fact that we can only vary the unknown function argument $x(t)$ while we can't vary there the independent variable $t$.
Functionals, variations, mechanics, virtual work at fixed time. This is the very same reason why if you have to evaluate virtual work (or use any other variation method in mechanics) you need to compute that work at "fixed time", i.e. without the variation of the independent variable time $t$.
Lagrangian and action
Sometimes the functional is written as an integral, like the relationship between the action $S$ and the Lagrangian function $L$,
\begin{equation}
S[x(t)] = \int_{t_1}^{t_2} L(\dot{x}(t), x(t), t) dt \ .
\end{equation}
If you proceed in the same way we did before, introducing a "small" variation $\varepsilon \eta(t)$ to the unknown function $x(t)$, and evaluate the incremental variation for $\varepsilon \rightarrow 0$, you end up with
\begin{equation}
\delta S = \int_{t_1}^{t_2} \left[ \dot{\eta}(t) \dfrac{\partial F}{\partial \dot{x}}(\dot{x}(t), x(t), t) + \eta(t) \dfrac{\partial F}{\partial x}(\dot{x}(t), x(t), t) \right] dt \ .
\end{equation}
Prescribing the stationarity of the functional $S$, i.e. $\delta S = 0$, and performing integration by part, and the arbitrarity of $\eta(t)$ except for the conditions $\eta(t_1) = \eta(t_2) = 0$, it's easy to get now Lagrange equations
\begin{equation}
\dfrac{d}{dt} \dfrac{\partial L}{\partial \dot{x}} - \dfrac{\partial L}{\partial x} = 0 \ .
\end{equation}