Matching in Blanchard-Diamond (1994)
The total number of matched applications is 

$$\sum_{x=1}^\infty g(x)xv = E[x]v$$

The number of applications matched with vacancies with $x$-many applications is 

$$g(x)xv$$

Hence 

$$Prob(\text{matching with x} | \text{matched} = \frac{g(x)x}{E[x]}$$

Using this in the equation in the question gives

$$ \tilde f = a \sum_{x=1}^\infty \frac{g(x)x}{E[x]}\frac{1}{x}\\
= E[x]a \sum_{x=1}^\infty g(x) \\
= \frac{v}{u} (1-g(0)) = f $$