It is easy to see that $\int \psi d\delta_t$ where $\delta_t$ is the Dirac mass at $t$ is $\psi(\{t\})$
So you are starting from a $T \in C([0,1])^{**}$ and you are attaching to it the function $t \mapsto T(\delta_t)$. This has absolutely nothing to do with Mauldin's result, so I'm not sure why you are mentioning it.
Now let $V \subset C([0,1])^*$ be the vector subspace spammed by dirac masses.
It is easy to see that $v = \sum a_i \delta_{x_i}$ is a finite linear combination of (distinct) Dirac masses, then $\Vert v \Vert = \sum |a_i|$.
Hence, pick any bounded function on $X$, the function that maps $\delta_{x_i}$ to $f(x_i)$ is continuous on $V$, of norm $\max |f(x)|$.
By the Hahn-Banach theorem it extend into an element of $C([0,1])^{**}$ of the same norm.
So this mean that $\psi_{|X}$ can be basically any bounded function on $[0,1]$. In particular it has no reason to be measurable and your second question simply don't make sense.