Since every $x \in C[a, b]$ is continuous on a compact set, it's bounded.
For any $x \in C[a, b]$, we have:
$$
|f(x)| = \left|\int_a^b x(t) x_0(t) \, dt \right| \le \int_a^b |x(t)| \cdot |x_0(t)| \, dt \le \|x\| \int_a^b |x_0(t)| \, dt
$$
Thus, $f$ is bounded and its norm satisfies:
$$
\|f\| \le \int_a^b |x_0(t)| \, dt \newcommand{sgn}{\operatorname{sgn}}
$$
In fact, equality holds. To see this, consider the sign function $\hat x(t) = \sgn(x_0(t))$. By Lusin's theorem, there is exists a sequence of functions $x_n \in C[a, b]$ such that $\|x_n\| \le 1$ and $x_n(t) \to \hat x(t)$ as $n \to \infty$ for every $t \in [a, b]$. By the dominated convergence theorem, we have:
\begin{align}
\lim_{n \to \infty} f(x_n) &= \lim_{n \to \infty} \int_a^b x_n(t) x_0(t) \, dt \\
&= \int_a^b \lim_{n \to \infty} x_n(t) x_0(t) \, dt \\
&= \int_a^b \sgn(x_0(t)) x_0(t) \, dt \\
&= \int_a^b |x_0(t)| \, dt
\end{align}
For the second linear functional you added, approximate the following function via Lusin's theorem in a similar manner:
$$
\hat x(t) = \begin{cases} 1 & \text{if } t \le \frac{a+b}{2} \\
-1 & \text{if } t > \frac{a+b}{2}
\end{cases}
$$
Since for all $x\in C^1[a,b]$ we have $|f(x)|=|x'(t_0)|\leq \|x\|$ it follows that $\|f\|\leq1$.
In order to prove that in fact $\|f\|=1$ we may assume $a\leq0\leq b$ and $t_0=0$. Consider the functions
$$x_n(t):={t\over 1 + n^2 t^2}\ .$$
Then
$$x_n'(t)={1-n^2 t^2\over (1+n^2 t^2)^2}\ ,$$
and it is easy to see that $|x_n'(t)|\leq x_n'(0)=1$ for all $t\in\Bbb R$. Furthermore we can deduce that $|x(t)|$ takes its maximum value ${1\over 2n}$ at $t=\pm{1\over n}$. It follows that for sufficiently large $n$ we have
$$\|x_n\|=1+{1\over 2n}\ ,$$
so that $f(x_n)=x_n'(0)=1$ implies
$$\lim_{n\to\infty}{|f(x_n)|\over\|x_n\|}=1\ ,$$
as claimed.
Best Answer
If $\|u\| = |f(x_0)| \neq 0$, then let $$y_0 = \frac{u(x_0) + i u(ix_0)}{\|u\|}x_0.$$ Note that $\|y_0\| = \|x_0\| = \|u\|$. Further, \begin{align*} f(y_0) &= \frac{u(x_0) + i u(ix_0)}{\|u\|} f(x_0) \\ &= \frac{(u(x_0) + i u(ix_0))(u(x_0) - i u(ix_0))}{\|u\|} \\ &= \frac{|u(x_0) - i u(ix_0)|^2}{\|u\|} \\ &= \frac{\|u\|^2}{\|u\|} = \|u\| \in \Bbb{R}. \end{align*} Therefore, $u(y_0) = \Re f(y_0) = \|u\|$, as required.