The Taylor series of a function $f(x)$ at a point $a$ is
$$f(a) +\frac{f'(a)}{1!}(x-a)+\frac{f''(a)}{2!}(x-a)^2+\cdots\tag 1$$
So how can, for a small $\varepsilon,$ the value of $f(x+\epsilon)$ be approximated by (and this is the correct way, presumably):
$$f(x+\varepsilon)=f(x)+f'(x)\varepsilon+f''(x)\varepsilon^2+\cdots\tag 2$$
as in this youtube:
?
Writing a series such as...
$$f(x) +\frac{f'(x)}{1!}(x)+\frac{f''(x)}{2!}(x)^2+\cdots$$
leaving the derivatives of $f(x)$ as a function of $x$ defeats the purpose of getting real coefficients of a polynomial, and it probably makes no sense.
And if the idea is that $a=x+\varepsilon$ for any $x$ the expansion would make sense with the derivatives being evaluated at $f(x+\varepsilon)$ as in
$$\begin{align} f(x+\varepsilon)&=\small{f(x+\varepsilon)+f'(x+\varepsilon)\left(x-(x+\varepsilon)\right)+f''(x+\varepsilon)(x-\left(x+\varepsilon)\right)^2+\cdots}\\ &=f(x+\varepsilon)+f'(x+\varepsilon)\left(\varepsilon\right)+f''(x+\varepsilon)\left(\varepsilon\right)^2+\cdots \end{align}$$
instead of $f(x)$ as in Eq (2).
Thanks to Jack M and Ethan Bolker, and so to close the loop, the equivalence between
$$f(\color{red}x)=f(\color{red}a) +\frac{f'(\color{red}a)}{1!}(\color{red}{x-a})+\frac{f''(\color{red}a)}{2!}(\color{red}{x-a})^2+\cdots\tag {1}$$
and
$$f(\color{blue}x+\color{blue}\varepsilon)=f(\color{blue}x)+\frac{f'(\color{blue}x)}{1!}\color{blue}\varepsilon+\frac{f''(\color{blue}x)}{2!}\color{blue}\varepsilon^2+\cdots\tag 2$$
is:
$$\begin{align} \color {blue} x &=\color {red} a\\ \color{blue} \varepsilon &= \color{red}{x-a} \end{align}$$
Hence, the Taylor expansion becomes:
$$f(x+\varepsilon)=f(\color{blue}x) +\frac{f'(\color{blue}x)}{1!}(\color{blue} \varepsilon)+\frac{f''(\color{blue}x)}{2!}(\color{blue} \varepsilon)^2+\cdots$$
Of note, there is a missing $\frac{1}{2!}$ in the original YT video for the third term of the Taylor expansion.