I'm not sure whether my answer is conceptual in your sense, but here is a relatively short proof. First of all, your definition of $f$ suggests the notation $$s_i s_p := \sum_{i=p}^n x_i.$$ Now consider the following telescopic sum: $$\label{eq} (1 - z_2) + z_2(1 - z_3) + z_2 z_3 (1 - z_4) + \dotsm + z_2 \dotsm z_{n-1} (1 - z_n) + z_2 \dotsm z_n = 1. \quad (*)$$ For each $i \in {2,\dots,n}$, take $$z_i = \frac{s_i}{x_1 + s_i},$$ hence $$1 - z_i = \frac{x_1}{x_1 + s_i},$$ and plug this into the telescopic sum $(*)$. Divide both sides of the equation by $x_1 \cdot s_2 s_3 \dotsm s_n$ to get the desired expression.
I'm not sure whether my answer is conceptual in your sense, but here is a relatively short proof. First of all, your definition of $f$ suggests the notation $$s_i := \sum_{i=p}^n x_i.$$ Now consider the following telescopic sum: $$\label{eq} (1 - z_2) + z_2(1 - z_3) + z_2 z_3 (1 - z_4) + \dotsm + z_2 \dotsm z_{n-1} (1 - z_n) + z_2 \dotsm z_n = 1. \quad (*)$$ For each $i \in {2,\dots,n}$, take $$z_i = \frac{s_i}{x_1 + s_i},$$ hence $$1 - z_i = \frac{x_1}{x_1 + s_i},$$ and plug this into the telescopic sum $(*)$. Divide both sides of the equation by $x_1 \cdot s_2 s_3 \dotsm s_n$ to get the desired expression.