In order to prove minimal sufficiency, a sufficient statistic $S(X)$ is needed to be considered. By the factorisation theorem, we can obtain functions g and h such that $$f_X(x; θ) = h(x)g(S(x), θ).$$ Therefore, if $S(x) = S(y)$ holds, then
$$\begin{align} f_X(x; θ) &= h(x)g(S(y), θ) \\&= h(x)/h(y)f_X(y, θ) \\&= k(x, y)f_X(y; θ). \end{align} $$
Characterisation of the minimal sufficiency is the following:
Let $X=\{X_{1},X_{2},...,X_{n}\}$ be a random sample from $f_{x}(x;\theta).$ If we can find a function $T$ such that $$\begin{align} T(x)&=T(y)\Longleftrightarrow \\f_{x}(x;\theta)&=k(x,y)f_{x}(y;\theta)\Longleftrightarrow\\\frac{f_{x}(x;\theta)}{f_{x}(y;\theta)}&=k(x,y),\end{align} $$ then $T(X)$ is a minimal sufficient statistic for $\theta.$ Here $k(x,y)$ does not depend on $\theta.$
By the conditions in the proposition, this implies that $T(x) = T(y).$ Hence we can find a function $R$ such that $T(x) = R(S(x))$ for all $x.$ Therefore, $T(X)$ is minimal sufficient.
The question is: Why $T(x) = R(S(x))$ for all $x$?