Definition of Strongly Parsimonious Reduction
A search problem is defined by a relation like $R$. The task is given an input $x$ find a solution $y$ s.t. $R(x,y)$ is true.

A counting problem is defined as counting the number of solutions of a search problem, i.e. given an input $x$ we want to compute the number of $y$s s.t. $R(x,y)$ holds. The counting problem associated with search problem $R$ is denoted by $\#R: \Sigma^* \to \mathbb{N}$, $x \mapsto \#y: R(x,y)$.

A parsimonious reduction $f$ between counting problems only needs to preserve the number of $y$s, i.e. for all $x$,

$$\#y:R(x,y) = \#z:S(f(x),z)$$

A strongly parsimonious reduction has an the extra property that you mentioned. The meaning of the property should become clear if you look at it in the following way:


  Assume that we want to use the search problem $S$ to solve the search problem $R$.
  On an input $x$ for $R$, we first compute $g(x)$ which is an input for $S$.
  Then we use $S$ to find an $S$-solution $y$ for $g(x)$, i.e. $(g(x),y)\in S$.
  Finally we compute $h(x,y)$ to obtain an $R$-solution for $x$.


Now if we want to preserve the number of solutions between these two search problems, we can make sure that for every $x$, the function $h(x,\cdot)$ that computes an $R$-solution for $x$ from an $S$-solution $y$ for $g(x)$ is a bijection.

The difference should be clear now: not only the number of $R$-solutions for $x$ and $S$-solutions for $g(x)$ are equal, there is an efficient bijection between those solutions.