Finding the norm of an operator.

Yes, your bound is correct. Let $X = C[0,1]$ with the $\sup$-norm and let $\{\alpha_k\}_{k = 1}^n \subset [0,1]$ as in the question. Let $f \in X$ be arbitrary with $\Vert f \Vert = 1$, then

$$ \vert Tf \vert = \vert \sum_{k = 1}^n \alpha_k f(x_k) \vert \leq \sum_{k = 1}^n \vert \alpha_k \Vert f \Vert_{\infty} \vert = \sum_{k = 1}^n \vert \alpha_k \vert $$

and hence $$\Vert T \Vert = \sup_{f \in X, \Vert f \Vert = 1} \vert Tf \vert \leq \sum_{k = 1}^n \vert \alpha_k \vert ~~.$$

In fact, this is the norm of the functional, since $\sum_{k = 1}^n \vert \alpha_k \vert$ is attained by the following function $f$. Define

$$ f(x) = \begin{cases} 1 & x = x_k, \alpha_k \geq 0 \\ -1 & x = x_k, \alpha_k < 0 \\ 0 & x \in \{0,1\} \setminus \{x_k\}_{k =1}^n \end{cases}$$

and linear interpolation otherwise. Then this function is continuous on $[0,1]$, it has $\Vert f \Vert = 1$, and we have

$$ \vert Tf \vert = \vert \sum_{k = 1}^n \alpha_k f(x_k) \vert = \sum_{k = 1}^n \vert \alpha_k \vert$$

where in the last step we use that all summands are positive. Hence $\Vert T \Vert \geq \sum_{k = 1}^n \vert \alpha_k \vert$, and thus we have equality.