It is important to realize that these are vectorial equations. The work is the integral of the force towards the path, along the path. The electric force created by the potential always points donwlope, and hence is always perpendicular to the equipotential surface. It thus has no component along the equipotential line, and thus contributes exactly zero to the integral.

It is important to realize that these are vectorial equations. The work is the integral of the force towards the path, along the path. More explicitly: your equation $W=Fs$ involves the scalar quantities "length of path" $s$ and "strength of force" $F$. It works well for a straight path, and a force in the same direction. But more generally, the path may be crooked and the force in any direction. So in each small segment of the path $\vec{ds}$ (vector!), the work along that segment is $\vec{F}\cdot\vec{ds}=F_{par}ds$ with $F_{par}$ being the component of the force in the direction of the path $\vec{ds}$. The work over the entire path is just the sum (integral) along the whole path, $$W=\int \vec{F}\cdot\vec{ds}$$.

The electric force created by the potential always points donwlope, and hence is always perpendicular to the equipotential surface. It thus has no component along the equipotential line, and thus contributes exactly zero to the integral.