If $c1$ and $c2$ are two paths with the same endpoints, then how come if $\int_{c1} F · ds = \int_{c2} F · ds$, then it's equivalent to $\int_c F · ds = 0$? $c$ is the path you get by taking $c1$ and then $c2$ backwards
Is it because they are the same distance so they cancel each other out?
The reason is because integrating along a curve backwards multiplies the integral by $-1$, that is, if $\gamma'$ is just the curve $\gamma$ in the opposite direction, $\int_\gamma F\cdot ds=-\int_{\gamma'} F\cdot ds$. Ultimately this is true because of the vector identity $u\cdot(-v)=-(u\cdot v)$. Thus when you integrate backwards along $c_2$, you get the negative of what you got when integrating along $c_1$, and it cancels out.
Breaking up $c$ into $c_1$ and $c_2'$ (which is $c_2$ backwards):
$$\int_c F\cdot ds=\int_{c_1} F\cdot ds+\int_{c_2'} F\cdot ds=\int_{c_1} F\cdot ds-\int_{c_2} F\cdot ds=0$$