The answer is negative. Indeed, for simplicity, let $a=-1$ and $b=1$. 

In the case when $\delta=1/10$, $n=1$, and $X_1$ is uniformly distributed on $[-1,1]$ (so that $\mu=0$), let $f(\mathbf X,\delta):=1-1/10$. 
Then 
$
\Pr\left (\left | \mu -\frac{1}{n} \sum_{i=1}^n X_i \right | \leq f(\mathbf X, \delta) \right ) =
\Pr(|X_1|\le1-1/10) = 1-1/10=1-\delta
$,
so that $(*)$ holds -- 
whereas 
$$
\Pr\left ( \mu -\frac{1}{n} \sum_{i=1}^n X_i  \leq \frac{1}{2}f(\mathbf X, \delta) \right )
=\Pr\left(X_1\ge-\frac12\,(1-1/10)\right)=29/40 
$$
$$\not\geq 1-1/10=1-\delta
,$$ 
so that $(**)$ fails to hold.  

In all cases other than the one just considered, let e.g. $f(\mathbf X, \delta) := b\sqrt{\frac{\ln(1/\delta)}{2n}}$, so that, by Hoeffding's inequality, $(*)$ hold.