Which reaction is more violent?
Although fluorine is more electronegative than chlorine it has less electron gain enthalpy because of its small size. 

The electrons in the outer shell of fluorine repel the incoming electron more effectively than the electrons in chlorine. More closer the electron comes to the nucleus more energy is released. In case of chlorine, the electron comes more close to the nucleus.

Thus, chlorine has more electron gain enthalpy.

Size matters in chemistry. It is more dominating factor than electronegativity in most cases. 

You would not expect enormous difference in enthalpy of formation of $\ce{CsF}$ and enthalpy of formation of other alkali halides. In case of $\ce{CsF}$, The bond length of the resultant salt will be larger than other alkali halides due to large size of cesium atom and small size of fluorine atom. The bond between them would not be very strong. Consequently, the reaction will not be strongly exothermic.

Putting the theory part aside, here is the data :

$\Delta{H^{\circ}_f}$  -  $\ce{CsF}$ $\rightarrow$ $-550\frac{KJ}{mole}$,

$\Delta{H^{\circ}_f}$ - $\ce{NaF}$ $\rightarrow$ $-575\frac{KJ}{mole}$,

$Electron\space gain\space enthalpy\space - Chlorine$ $\rightarrow$ $-349 \frac{KJ}{mole}$

The comparisons in your question are very different and they can only be compared by looking at the data. The prediction will be very inaccurate. If you want to compare two properties, you will at least need some similarity between the two properties. Like, you can compare enthalpy of formation of two different alkali halides or electron gain enthalpies of Halogens or chalcogens. Electron gain enthalpy and enthalpy of formation of alkali halides are very different things.