Just start from the top with how they applied the chain rule.
What have I tried:
On
First of all, they are not differentiating $f$, they are differentiating the function $g(t)=f(p+t(x-p))$. The function $g$ is the composition of the function $f$ with the function $h(t)=p+t(x-p)$ (this is a function $\mathbb{R}\to\mathbb{R}^n$). So by the chain rule, $$\frac{dg}{dt}(t)=\sum \frac{dh^i}{dt}(t)\frac{\partial f}{\partial x_i}(h(t)),$$ where $h^i$ is the $i$th component of $h$. Since $h^i(t)=p^i+t(x^i-p^i)$, $\frac{dh^i}{dt}$ is just $x^i-p^i$, so this gives the formula shown.
I don't know if this helps. Observe \begin{align} f(x_1, \ldots, x_n) \end{align} and \begin{align} x_i = p_i+t(x_i-p_i) \end{align} then by chain rule, we have \begin{align} &\frac{d}{dt}f(p_1+t(x_1-p_1), p_2+t(x_2-p_2), \ldots, p_n+t(x_n-p_n)) \\ =& \frac{\partial f}{\partial x_1} \frac{d x_1}{d t}+ \frac{\partial f}{\partial x_2} \frac{dx_2}{dt}+\ldots + \frac{\partial f}{\partial x_n}\frac{dx_n}{dt}\\ =&\ \frac{\partial f}{\partial x_1} (x_1-p_1) + \frac{\partial f}{\partial x_2}(x_2-p_2) + \ldots +\frac{\partial f}{\partial x_n}(x_n-p_n)\\ =&\ \nabla f(p+t(x-p)) \cdot (x-p). \end{align}