$f$ : $U(\subset \mathbb{R}^n) \to \mathbb{R}$, $x_0 \in U^i$
We fix $x \in U$ and we examine $f(x_0 + t(x-x_0))$ .
$\left (f(x_0 + t(x-x_0)) \text{ is a function of } "t". \right)$
Then, my textbook says that
\begin{align} &\dfrac{d}{dt} f(x_0 +t(x-x_0))\\= &\sum_{i=1}^n \dfrac{\partial f}{\partial x_i} (x_0+t(x-x_0))(x-x_0)_i.\\ &(x-x_0)_i \text{ means $i$-th component of } x-x_0. \end{align}
I cannot understand why this holds.
I think that this is related to chain rule, but I cannot understand.(For example, why do I have to sum from $i=1$ to $i=n$ , and why $(x-x_0)_i$ appears.)
Because when you write $f$ differential at $x_0$ using the partial derivatives you have
$$f^\prime(x_0).h =\sum_{i=1}^n \frac{\partial f}{\partial x_i}h_i$$
Where $h_i$ stands for the $i$-th coordinate of $h$. Then as you suspected, it is an application of the chain rule as the differential of $$t \mapsto x_0+t(x-x_0)$$ is $x-x_0$.