The quick and general answer is electrostatic repulsion. Fructose-6-phosphate is already a dianion. Phosphorylating it a second time will mean adding another two negative charges to the dianion. Like charges repel each other, so the two dianionic phosphate groups are pushed apart electrostatically.
However, you cannot generalise this. Sometimes, the additional negative charge may serve to better balance charges overall. It could also be so far away from an existing negative charge that electrostatic interactions are minimal. Finally, there are many ions and positively charged organic molecules dissolved in the cell which altogether stabilise negative charges relatively with respect to a simple solution of fructose-1,6-bisphosphate. (The same can be said for the stabilisation of positive charges.)
Finally, please note that the additional negative charge does not really assist the fragmentation of fructose-1,6-bisphosphate. The reaction — a retro-aldol — is enzymatic and catalysed principally by a lysine and aspartate residue. The former is used to convert the keto functionality into an imine which has a greater tendency to form an enamine while the latter serves as a proton shuttle. The phosphate groups are merely there so the enzyme recognises the correct molecule; modifying the enzyme’s structure to allow for unphosphorylated fructose to be consumed would not alter the reaction rate significantly.