Fisher's factorisation theorem is one of several ways to establish or prove that a statistic $S_n(X_1,\ldots,X_n)$ is sufficient. The meaning of sufficiency remains identical through all these manners of characterising it though, namely that the conditional distribution of the sample $X_1,\ldots,X_n$ conditional on $S_n(X_1,\ldots,X_n)$ is constant in $\theta$, i.e., is the same distribution for all values of $\theta$.
For instance, given an iid sample $X_1,\ldots,X_n$, the order
statistic $S_n(X_1,\ldots,X_n)=(X_{(1)},\ldots,X_{(n)})$ is sufficient
because the distribution of $X_1,\ldots,X_n$ given
$S_n(X_1,\ldots,X_n)$ is the uniform distribution on the permutations
of $\{1,\ldots,n\}$: \begin{align*}\mathbb
P_\theta((X_1,\ldots,X_n)&=(y_1,\ldots,y_n)|S_n(X_1,\ldots,X_n)=(x_{(1)},\ldots,x_{(n)})\\&=\frac{\mathbb
I_{(x_{(1)},\ldots,x_{(n)})}(y_{(1)},\ldots,y_{(n)})}{n!}\end{align*}
The factorisation theorem thus does not modify the original distribution of the sample and does not particularly help in finding it. It operates the other way, as a mean of figuring out sufficient statistics by looking at the joint pmf or pdf. Once a sufficient statistic has been found, its own distribution is sufficient (in the sense of "enough") for creating a likelihood function in the parameter $\theta$. It will then differ from the "full" likelihood by a constant (in the parameter) equal to the function (of the sample) $h(x_1,\ldots,x_n)$ found in the factorisation theorem. But this constant has no relevance for statistical inference.
