Friedel craft alkylation halo-carbon bond cleavage
In Friedel-Crafts alkylation, the halogen-carbon bond breaking depends on the strength of bond between halogen atom and the Lewis acid used. The more stronger is the bond, the easier it is to generate a carbocation for the reaction.( In all cases, the carbocation exoplicitly needn't have to generate, a strong $\delta$+ charge on the electrophilic carbon is sufficient enough to have a $\ce{S_E2}$ or $\ce{S_E1}$ type reaction. Here, the Lewis acid used is  $\ce{BCl_3}$ where there is a vacant  $\ce{2p_z}$ orbital on the  $\ce{B}$ atom,(Assuming, the molecule is oriented in a triangle in the  $\ce{xy}$-plane.)and the Lewis base is the given reagent 1-chloro-3-fluoropropane., which has two Basic centre. But donation through  $\ce{F}$ will correspond to a stronger  $\ce{2p$\pi$ -2p$\pi$}$ overlap with Lewis acid,rather than weaker  $\ce{3p$\pi$ -2p$\pi$}$ overlap in the case of donation through  $\ce{Cl }$ atom. Thus bonding with  $\ce{F}$ will correspond to a very huge amount of  $\ce{-I}$ effect of the adjacent carbon atom which will weaken the bond between them by pulling the electron pair towards  $\ce{F}$ and easily favour the attack by the $\pi$-electron cloud of Benzene ring.Also, the repulsion between $\pi$-electron cloud and the  large sized $\ce{Cl}$ atom is very less during the formation of  the above compound. But, if it would have bonded through  $\ce{Cl}$, there won't be a great electrophilic nature of the adjacent carbon due to less effective bonding in the case of  $\ce{Cl}$(corresponding to even lesser  $\ce{-I }$ effect),so attacking will correspond to a highly energetic transition state, and also, there will be a slight electrostatic repulsion with the  $\ce{F}$ atom of the  $\ce{$\pi$ }$-electron cloud. 
Due to such above mentioned phenomena and energetically favoured conditions, bonding through cleavage of  $\ce{C-F}$ bond is more preffered over cleavage of  $\ce{C-Cl}$ bond.