The existence of the derivative of a Banach space valued function In Evans' Partial Differential Equations $\S 7.1$, there is a motivation for definition of weak solution.

\begin{align*} 
(1) \begin{cases} u_t + Lu = f \ &\text{in} \ U_T\\
 u = 0 \ &\text{on} \ \partial U \times [0,T]\\
u = g \ &\text{on} \ U \times \{ t = 0 \}  \end{cases}.
\end{align*}


$\textbf{Motivation for definition of weak solution.}$ To make plausible the following definition of weak solution, let us first temporarily suppose that $u = u(x,t)$ is in fact a smooth solution in parabolic setting of our problem $(1)$. We now switch our viewpoint, by associating with $u$ a mapping


$$\textbf{u}: [0,T] \longrightarrow H^1_0(U)$$


defined by


$$[\textbf{u}(t)](x) := u(x,t) \ (x \in U, 0 \leq t \leq T).$$


In other words, we are going consider $u$ not as a function of $x$ and $t$ together, but rather as a mapping  $\textbf{u}$ of $t$  into the space $H^1_0(U)$ of functions of $x$. This point of view will greatly clarify the following presentation.


Returning to the problem $(1)$, let us similary define


$$\textbf{f}: [0,T] \longrightarrow L^2(U)$$


by


$$[\textbf{f}(t)](x) := f(x,t) \ (x \in U, 0 \leq t \leq T).$$


Then if we fix a function $v \in H^1_0(U)$, we can multiply the PDE $\frac{\partial u}{\partial t} + Lu = f$ by $v$ and integrate by parts, to find


$$ (9)\ (\textbf{u}',v) + B[\textbf{u},v;t] = (\textbf{f},v) \ \left( ' = \frac{d}{dt} \right)$$


for each $0 \leq t \leq T$, the pairing $(,)$ denoting the inner product of $L^2(U)$.

If we multiply the PDE $\frac{\partial u}{\partial t} + Lu = f$ by $v \in H^1_0(U)$, then we would get $(u_t,v) + B[\textbf{u},v;t] = (\textbf{f},v)$. However, the first term in (9) is $(\textbf{u}',v)$ instead of $(u_t,v)$.
As far as I know, $\textbf{u}' = \lim\limits_{h \to 0}\frac{\textbf{u}(t+h) - \textbf{u}(t)}{h}$ in $H^1_0(U)$, in other words, $ \Big\Vert \lim\limits_{h \to 0}\frac{\textbf{u}(t+h) - \textbf{u}(t)}{h} - \textbf{u}' \Big\Vert_{H^1_0(U)} \to 0$. On the other hands, $u_t = \lim\limits_{h \to 0} \frac{u(t+h,x) - u(t)}{h}$ and the limit is pointwise.
So, I am wondering how to show that $\textbf{u}'$ exists from the existence of $u_t$ and also $\textbf{u}'$ is equal to $u_t$. Any help would be very much appreciated!

In the context of the book, $\mathbf{u}'$ isn't the classical derivative as defined in your post.
As defined in Section 5.9.2 (Spaces involving time), $\mathbf{u}'$ stands for the "weak" derivative of $\mathbf{u}$, that is, a function $\mathbf{v}$ such that
$$\int_0^T \varphi'(t)\mathbf{u}(t)\;dt=-\int_0^T \varphi(t)\mathbf{v}(t)\;dt,\quad\forall \ \varphi\in C_c^\infty(0,T).$$
Since $\mathbf{v}=u_t$ satisfies the above equality, we have $\mathbf{u}'=u_t$ in the weak sence (which is the book's claim).
