In this video, the instructor uses Undetermined Coefficients (or at least what would be called Undetermined Coefficients in the ODE world) to find a particular solution to the PDE $U_t + cU_x = f(x + ct)$, where $c$ is a parameter present in the original Wave Equation.
The guess used is $U = aF(x + ct)$, where $a$ is the constant to be determined, and $F$ is said to be the antiderivative of $f$. However, I don't recall from MV Calc any single notion of an antiderivative of a multivariable function, only antiderivatives taken with respect to one of the function's inputs. The closest thing I can think of would be a potential function, where one begins with a vector field and pieces together a single function whose gradient is that vector field, but I don't think that matches what's going on here.
Later in the problem, he seems to take $\frac{\partial F}{\partial x}$ and $\frac{\partial F}{\partial t}$ separately, in both cases yielding $f$, but it seems to me that baring certain exceptional cases, it is not possible to find a function whose partial with respect to each of its variables yield a single, given function. What is going on here?
If $f$ is a function of a single variable and $s$ is a function of two variables $x$ and $t$ the composition $f \circ s$ is also a function of $x$ and $t$. This is often written in a convenient but slightly imprecise form as $$f(x,t) = f(s(x,t)).$$ In your case $f$ is a function of a single variable and its antiderivative in the usual sense is $F$. The function $s$ is defined as $s(x,t) = x + ct$.
Using the same notation you may write $$F(x,t) = F(s(x,t)) = F(x + ct).$$ The chain rule states $$\frac{\partial F}{\partial x} (x,t) = F'(s(x,t))\frac{\partial s}{\partial x}(x,t) = f(s(x,t))$$ and similarly $$\frac{\partial F}{\partial t} (x,t) = F'(s(x,t))\frac{\partial s}{\partial t}(x,t) = c f(s(x,t))$$ giving you $$c \frac{\partial F}{\partial x}(x,t) = \frac{\partial F}{\partial t}(x,t).$$