We know that Taylor's series expansion of a generic function $f(x)$, in the abscissa point $x_0\in\mathbb{R}$ is given by:
$$f(x)=\sum_{n=0}^{\infty}\frac{(x-x_0)^n}{n!}f^{(n)}(x_0) \tag{1}$$
$$f(x)=\left[e^{(x-x_0)\frac{\partial}{\partial x}}\right]f(x_0) \tag{2}$$ Why if I expand the exponential in $(2)$ into an infinite sum and move the $(x)$ part into the sum, I'll get $(1)$ through a translation that brings the point $x_0$ to the point $x=x_0+(x-x_0)$ where $$e^{(x-x_0)\frac{\partial}{\partial x}}\equiv \sum_{n=0}^{\infty}\frac{(x-x_0)}{n!}\frac{\partial^{(n)}}{\partial x^n}, \quad ?$$
What is the connection between moment operator impulse $p_x\longrightarrow -i\hbar\frac{\partial}{\partial x}$ along a direction, $x$ and the $(3)$?
$$f(x)=\left[e^{(x-x_0)\frac{\partial}{\partial x}}\right]f(x_0)\color{teal}{\equiv \left[e^{-\dfrac i\hbar (x-x_0)p_x}\right]} \tag{3}$$
Let's first start with a precise formulation of the equation you are referring to. For a (sufficiently nice) function $f$ we have $$f(x+\epsilon) = e^{\epsilon\frac{d}{dx}}f(x)$$ Note that we need it to act on a function $f(x)$ and not $f(x_0)$ as you have (as this is just a constant). Also we cannot have $(x-x_0)\frac{d}{dx}$ in the exponential since $((x-x_0)\frac{d}{dx})^n \not= (x-x_0)^n\frac{d^n}{dx^n}$.
As to why this equation holds. First of all the definition of $e^A$ when $A$ is an operator is nothing but the usual Taylor series so $e^A \equiv 1 + A +\frac{A^2}{2} + \ldots$ so if we write out what we mean by the right hand side it becomes the series $$(1 + \epsilon \frac{d}{dx} + \frac{\epsilon^2}{2}\frac{d^2}{dx^2} + \ldots)f(x)\\ = f(x) + \epsilon \frac{df(x)}{dx} + \frac{\epsilon^2}{2}\frac{d^2f(x)}{dx^2}+\ldots$$ If you consider this as representing a function of just $\epsilon$ for a fixed $x$ then you will see that this is nothing but the Taylor expansion of the function $g(\epsilon) = f(x+\epsilon)$ about the point $\epsilon=0$.
The connection with the momentum operator becomes $f(x+\epsilon) = e^{-\frac{i}{\hbar} \epsilon p_x }f(x)$. This is sometimes phrased as the momentum operator being the generator for translations (which intuitively makes sense: momentum is what causes change in position). To understand this more deeply I suggest reading more about the mathematics of quantum mechanics which can be found in any introductory textbook. For more see e.g. this or this.