Radon–Nikodym Derivative and Bayes' Theorem

You wrote: $$ \frac{\mathrm d\mu_{\Theta\mid X}}{\mathrm d\mu_\Theta}(\theta \mid x) = \frac{f_{X\mid \Theta}(x\mid \theta)}{\int_\Omega f_{X\mid\Theta}(x\mid t) \, \mathrm d\mu_\Theta(t)} $$ Let's rearrange it a little bit: $$ \mathrm d\mu_{\Theta\mid X} (\theta \mid x) = \frac{f_{X\mid \Theta}(x\mid \theta) \, \mathrm d\mu_\Theta}{\int_\Omega f_{X\mid\Theta}(x\mid t) \, \mathrm d\mu_\Theta(t)} $$ and then: $$ \frac{\mathrm d\mu_{\Theta\mid X}}{d\nu} (\theta \mid x) = \frac{f_{X\mid \Theta}(x\mid \theta) \, (\mathrm d\mu_\Theta/d\nu)(\theta)}{\int_\Omega f_{X\mid\Theta}(x\mid t) \, \mathrm d\mu_\Theta(t)} $$ $${}$$ $$ \frac{d\mu_{\Theta\,\mid\, X=x}}{d\nu}(\theta) = \frac{ \displaystyle \frac{d\mu_{X\,\mid\,\Theta=t}}{d\lambda}(x) \cdot \frac{d\mu_\Theta}{d\nu}(\theta) }{ \displaystyle \int \frac{d\mu_{X\,\mid\,\Theta=t} (x)}{d\lambda} \cdot d\mu_\Theta(t) } $$

You seem to be confused about how to reconcile the familiar version of Bayes' theorem $$ \tag{1}\label{1} p(\theta \mid x) = \frac{p(\theta) p(x \mid \theta)}{p(x)} $$ with the formal version presented here: $$ \tag{2} \label{2} \frac{d\mu_{\Theta\mid X}}{d\mu_\Theta}(\theta \mid x) = \frac{f_{X\mid \Theta}(x\mid \theta)}{\int_\Omega f_{X\mid\Theta}(x\mid t) \, d\mu_\Theta(t)}. $$ (I will be using the same notation as at that link.)

On the one hand, the left-hand-side of \eqref{1} is supposed to represent a density of the conditional distribution of $\Theta$ given $X$ with respect to some unspecified dominating measure on the parameter space.

On the other hand, the left-hand-side of \eqref{2} is the density of the conditional distribution of $\Theta$ given $X$ with respect to the prior distribution.

If, in addition, the prior distribution $\mu_\Theta$ has a density $f_\Theta$ with respect to some (let's say $\sigma$-finite) measure $\lambda$ on the parameter space $\Omega$, then $\mu_{\Theta \mid X}(\cdot\mid x)$ is also absolutely continuous with respect to $\lambda$ for $\mu_X$-a.e. $x \in \mathcal{X}$, and if $f_{\Theta \mid X}$ represents a version of the Radon-Nikodym derivative of $d\mu_{\Theta\mid X}/d\lambda$, then \eqref{2} yields $$ \begin{aligned} f_{\Theta \mid X}(\theta \mid x) &= \frac{d \mu_{\Theta \mid X}}{d\lambda}(\theta \mid x) \\ &= \frac{d \mu_{\Theta \mid X}}{d\mu_\Theta}(\theta \mid x) \frac{d \mu_{\Theta}}{d\lambda}(\theta) \\ &= \frac{d \mu_{\Theta \mid X}}{d\mu_\Theta}(\theta \mid x) f_\Theta(\theta) \\ &= \frac{f_\Theta(\theta) f_{X\mid \Theta}(x\mid \theta)}{\int_\Omega f_{X\mid\Theta}(x\mid t) \, d\mu_\Theta(t)} \\ &= \frac{f_\Theta(\theta) f_{X\mid \Theta}(x\mid \theta)}{\int_\Omega f_\Theta(t) f_{X\mid\Theta}(x\mid t) \, d\lambda(t)}. \end{aligned} $$

Hopefully this shows you how to get to the familiar form \eqref{1} from \eqref{2}.