Equation 20 doesn't have a $w$ in it, and the first appearance of $w$ is in equation 20a, so it's not a simple deduction from previous lines. But a particular example might be helpful.
So take the Schwarzschild solution, and fix two points $(t_1,x_1,y_1,z_1)$ and $(t_2,x_2,y_2,z_2)$ outside the event horizon. Assume there is a geodesic between them that is everywhere timelike (so the tangent to the geodesic is timelike at every point on the geodesic). This might not always hold, but let's do it for this example. Now there are otehr paths, including paths where $t$ stays the same, but in a sense they are "far" from the actual path, so for the close paths we can parameterize the path by the coordinate $t$ and compute
$$\int_{(t_1,x_1,y_1,z_1)}^{(t_2,x_2,y_2,z_2)}ds=\int_{t_1}^{t_2}\frac{ds}{dt} dt$$ for any path where $s$ is a cumulative measure of arclength. It's really just a sum over a finite number of segments of the arc length. What is the arclength for a little segment of the curve? It's just the square root of the squared length of the tangent vector (assuming the curve is small enough that the tangent approximates it well, and since this is an integral, the finite sum of the integral does evaluate this arbitrarily close). What is the tangent? Its how the curve moves as we change the $t$ coordinate a bit, so it has coordinates $\frac{\partial x ^\mu}{\partial t}$. How do we compute the squared length of such a vector $\frac{\partial x ^\mu}{\partial t}$? It's $$\Sigma_\mu\Sigma_\nu g_{\mu\nu}\frac{\partial x ^\mu}{\partial t}\frac{\partial x ^\nu}{\partial t}.$$
now let's call give the quantity $\Sigma_\mu\Sigma_\nu g_{\mu\nu}\frac{\partial x ^\mu}{\partial t}\frac{\partial x ^\nu}{\partial t}$ a name:
$$w^2=\Sigma_\mu\Sigma_\nu g_{\mu\nu}\frac{\partial x ^\mu}{\partial t}\frac{\partial x ^\nu}{\partial t}.$$
That's the squared length, so the length is $w$, so
$\int_{t_1}^{t_2}wdt$ is literally the arc length of the curve from one point to the other for a random path that is "close" to the geodesic. It's variation then is $\delta\int_{t_1}^{t_2}wdt=\int_{t_1}^{t_2}\delta wdt$.
This is exactly the same as equation 20 except that we parameterized by $t$. Now, let's generalize this example. The family of surfaces might not be a $t=const$ hypersurface, but as long as there is one point on the surface for each part of the curve, for every curve close to the geodesic then we are OK, and we can just fix that family and it doesn't matter. So really this comes from picking up on what $w$ is defined to be.
The next line after 20a come from implicit differentiation of $$w^2=\Sigma_\mu\Sigma_\nu g_{\mu\nu}\frac{\partial x ^\mu}{\partial t}\frac{\partial x ^\nu}{\partial t}.$$
As for equation 20b, again it is just a definition of $\kappa$ that you use to show the previous line.
edit to explain the minus sign in 20b
Maybe it's more clear if you multiply all of line 20b by a negative sign (doesn't change the real result which is the line before 20b). Because then looking at $$\delta w=\Sigma_\sigma\Sigma_\mu\Sigma_\nu \frac{1}{2w}\frac{\partial g_{\mu\nu}}{\partial x^\sigma}\frac{d x ^\mu}{d t}\frac{d x ^\nu}{d t}\delta x^\sigma+\Sigma_\mu\Sigma_\nu \frac{1}{w}g_{\mu\nu}\frac{d x ^\mu}{d t}\delta\frac{d x ^\nu}{d t},$$ the first term $\Sigma_\sigma\Sigma_\mu\Sigma_\nu \frac{1}{2w}\frac{\partial g_{\mu\nu}}{\partial x^\sigma}\frac{d x ^\mu}{d t}\frac{d x ^\nu}{d t}\delta x^\sigma$ term doesn't change and the other term, $\Sigma_\mu\Sigma_\nu \frac{1}{w}g_{\mu\nu}\frac{d x ^\mu}{d t}\delta\frac{d x ^\nu}{d t}$, picks up a minus sign when you do integration by parts.
