How can a bond be more stable and more polar at the same time?
The important thing here to keep in mind is that the bond strength is not a spherical chicken in vacuum, meaning that we need to keep the entire process in mind when analyzing all the factors that can influence it. Although the fluorine has a higher electronegativity, it is significantly smaller than Cl, Br, or I (same goes to their corresponding ions). And without going too deep into intricacies, $\ce{F-}$ is $way$ less stable than $\ce{I-}$ and thus, using the OChem language, a much better nucleophile. Due to it's inability to stabilize the charge, it holds onto the hydrogen with all it's might (pardon me for anthropomorphizing the atoms).

SO, when you're thinking about breaking $\ce{H-F}$ bond, you gotta take into consideration what happens with the products. In the example with organic chemistry and leaving groups that you have pointed out, fluorine is the worst leaving group for the precisely the same reason--it cannot stabilize the negative charge well, so it doesn't "want" to leave. However, if you try to do a, say, $S_N2$ reaction in a protic solution (not the best way to go around, but doable), then $\ce{F-}$ will be a much better leaving group than $\ce{I-}$, b/c $\ce{F-}$ will instantly "catch" a hydrogen from the solution and stop being nucleophilic, which according to Le'Chatelie'r principle shifts the equilibrium to where $\ce{F-}$ is a leaving group.