Local boundedness will not help either.
Take $\mathbb{W} = \mathbb{N}$, which is a closed subset of $\mathbb{R}$ (all its points are isolated). Define $\mu$ by $\mu(\{n\}) = 2^{-n}$; this is a probability measure. Take $\mathbb{X} = \{1, \frac{1}{2}, \frac{1}{3}, \dots, 0\}$; this is a compact set. Note that all the points of $\mathbb{X}$ except 0 are isolated points.
For each integer $n$, let $f(1/n, n) = n 2^n$. Let $f(x,w) = 0$ for all other $x,w$. Now $f$ is (jointly!) continuous on $\mathbb{X} \times \mathbb{W}$. I want to emphasize that no smoothing or other modification to $f$ is needed to obtain this! $f$ is automatically continuous at points of the form $(1/n, m)$ since those points are isolated in $\mathbb{X} \times \mathbb{W}$. At points of the form $(0,m)$, we have $f(0,m)=0$. Now $U = \{(1/n, m) : n > m\} \cup \{(0,m)\}$ is an open neighborhood of $(0,m)$ in $\mathbb{X} \times \mathbb{W}$, and $f=0$ identically on $U$, so $f$ is continuous at $(0,m)$ also.
Being continuous, $f$ is automatically locally bounded, i.e. bounded on compact subsets of $\mathbb{X} \times \mathbb{W}$. Furthermore, $f(x, \cdot)$ is integrable for every $x$ (indeed, for every $x$, $f(x,\cdot)$ is a bounded function). We have $\bar{f}(1/n) = n$, and $\bar{f}(0) = 0$, so $\bar{f}$ is unbounded and the estimate you want cannot hold.
Fundamentally, the problem is that your conditions are ensuring that as $x_n \to x$, we have $f(x_n, \cdot) \to f(x, \cdot)$ pointwise, or if you require joint continuity, $f(x_n, \cdot) \to f(x, \cdot)$ locally uniformly. But what you need is something more like $f(x_n, \cdot) \to f(x,\cdot)$ in $L^1(\mathbb{W}, \mu)$ and that is a very different topology from the pointwise or compact-open topologies. To get it, you will need something like uniform integrability, and this will require some sort of global or uniform control.