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I am having trouble understanding a step in the following proof from "Metric Structures in Differential Geometry", part of which is attached below.

enter image description here

Here $M_p$ is the tangent space to $M$ at $p$ and $E_p$ is just a fiber of some vector bundle. $X$ is a section of $E$ along $f$. My question is how the author is able to conclude that $\nabla_{D_1(0)}\nabla_{D_2}X=0$ to arrive at the last line.

The justification is that $X$ is the parallel translation of $x$ from $p$ to $f(t,0)$ and then from $f(t,0)$ to $f(t,s)$ which is exactly the concatenation of the curves $(1)$ and $(2)$ whose initial tangent vectors are $D_1(0)$ and $D_2(t,0)$ respectively. Even though this is the case I can't quite join the dots to see why $\nabla_{D_1(0)}\nabla_{D_2}X=0$.

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    The section $X$ is defined as parallel transport of $x$ along the first two segments therefore by definition, its covariant derivative with respect to the velocity vectors are zero. – 2017-02-09
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    @tessellation I can see why we would have $\nabla_{D_1(0)}X=0$ since $X$ is parallel with respect to $D_1(0)$ but why is $\nabla_{D_2}X$ as a vector field also parallel with respect to $D_1(0)$? – 2017-02-09
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    Actually as $X$ is parallel transport along the first two curve the vector field $\nabla_{D_2}X$ is a zero vector field along the curve. – 2017-02-09
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    @tessellation Are you saying that $\nabla_{D_2}X=0$ along the first curve? – 2017-02-09
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    The definition of $X$ is the source of your confusion, so let me explain it to you. First transport $x$ along curve (1). Now for fixed $t=t_0$ you have a section at $f(0,t_0)$. Now transport this along $f(s,t_0)$. This way you get the section $X$. Now from definition it follows that for every $t$, $\nabla_{D_2}X=0$. Hence the first term is zero. – 2017-02-09
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    @tessellation Let me summarise your argument to see if it's right. At each point in time $t$ along the first curve we get a tangent vector $\nabla_{D_2}X(t)$ which is always $0$. So then $\nabla_{D_1}\nabla_{D_2}X(t)$ is also always $0$. Is this right? – 2017-02-09
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    Yes you are right. – 2017-02-09

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