The theorem in question is:
Now, the part that I have a problem is the following highlighted text:
Theorem (5.10) is
I don't understand what function plays the role of $f$ at (5.10). What i've tried is the function $$g(t)=f(x+v_{j-1}+th_je_j)$$ But its derivative depends on the deriative of $f$ which I don't know it exists yet.
Your idea of looking at the map $g(t) := f(x+v_{j-1}+th_{j}e_{j})$, which is defined on some open subset containing $[0,1]$, namely $\{t\in \mathbb{R} : x+v_{j-1}+th_{j}e_{j} \in E \}$, is indeed a good idea. To establish that $g$ is differentiable on $[0,1]$, note that $g$ is a composition of the maps $\phi_{1}$ and $\phi_{2}$ (defined on suitable open subsets) given by $\phi_{1}(t) := th_{j}$ and $\phi_{2}(s) := f(x+v_{j-1}+se_{j})$. The derivative of $\phi_{1}$ at $t$ is $h_{j}$ and the derivative of $\phi_{2}$ at $s$ is $(D_{j}f)(x+v_{j-1}+se_{j})$, which can be more clearly observed from
\begin{align*} \lim_{\xi\to 0}\frac{\phi_{2}(s+\xi ) - \phi_{2}(s)}{\xi}&= \lim_{\xi \to 0}\frac{f(x+v_{j-1}+(s+\xi )e_{j}) - f(x+v_{j-1}+se_{j})}{\xi} \\ &= \lim_{\xi \to 0}\frac{f((x+v_{j-1}+se_{j})+\xi e_{j}) - f(x+v_{j-1}+se_{j})}{\xi} \\ &= (D_{j}f)(x+v_{j-1} + se_{j}). \end{align*}
By the chain rule, the derivative of $g = \phi_{2} \circ \phi_{1}$ at $t$ is $h_{j}(D_{j}f)(x+v_{j-1}+th_{j}e_{j})$.
As $g$ is differentiable on $[0,1]$, it follows from the real-valued intermediate value theorem that there is some $\theta_{j} \in (0,1)$ such that $g(1) - g(0) = (1-0)g'(\theta_{j} ) = g'(\theta_{j} )$. As $g(0) = f(x+v_{j-1})$ and $g(1) = f(x + v_{j})$, it follows that there is some $\theta_{j} \in (0,1)$ such that
$$f(x+v_{j}) - f(x+v_{j-1}) = h_{j}(D_{j}f)(x+v_{j-1}+\theta_{j} h_{j}e_{j}).$$