Mechanism of acetylene formation from iodoform and silver
What happens is that the C-I bond being weak, the iodide group is attacked by the silver atom (similar to what Na does in Wurtz reaction). As, there are 3 I-atoms and 2 molecules of iodoform usually react, 6 silver atoms are used up in the reaction and they precipitate out 6 molecules of AgI. The 2 carbon atoms form a triple bond to form acetylene. 

EDIT- $$\ce{2CHI3 + 6Ag-&gt; C2H2 + 6AgI}$$
$$\ce{C-I + 2Ag +I-C-&gt;I\bond{...}Ag + Ag\bond{...}I + C\bond{...}C}$$
As there are 3 C-I bonds, 6 Ag atoms form bonds with I-atoms, Hence, triple bond is formed between the 2 C-atoms. I hope that explains it.