From Wald's General Relativity page 37,
It is easiest to compute the change in $v^a$ when we return to $p$ by letting $\omega_a$ be an arbitray dual vector field and finding the change in the scalar $v^a\omega_a$ as we traverse the loop.
1) Where is the vector $v^a$ in the picture above?
2) Why do they consider $\omega_a$?
For small $\Delta t$ the change, $\delta_1$, in $v^a\omega_a$ on the first leg of the path, $$\delta_1 = \Delta t \frac{\partial}{\partial t}(v^a\omega_a)|_{(\Delta t/2,0)}$$
where, by evaluating the derivative at the midpoint, we have ensured that this expression is accurate to second order in $\Delta t$. We may rewrite $\delta_1$ as $$ \delta_1 = \Delta t T^b \triangledown_b (v^a\omega_a)|_{(\Delta t/2,0)}$$
3) How can we introduce the $T^b \triangledown_b$?? I do not see where this came from?
It is just a convenient step, you will see how it pays off later in the development. At this point, just note that by introducing $\omega_a$, we can build the quantity $v^a\omega_a$, which is a scalar, and easier to work with: one less index to worry about
The chain rule
$$ \frac{\partial }{\partial t} = \left(\frac{\partial x^b(t, s)}{\partial t}\right)_s\frac{\partial}{\partial x^b} = T^b \partial_b $$
where
$$ T^b \stackrel{.}{=} \left(\frac{\partial x^b(t, s)}{\partial t}\right)_s $$
is the tangent to the curves of constant $s$