MTW's Gravitation, Box 11.2, etc.: Relative acceleration depends only on $\mathbf{u}$ and $\mathbf{n}$ at the fiducial point? My question relates to MTW's Gravitation, Box 11.2 (copied below) and the discussion on page 271.  This is my paraphrase of the gist of box 11.2:
Consider a family $\Lambda$ of timelike geodesics affinely parameterized
by $\lambda$ and selected by the parameter $n$. The fiducial geodesic
is designated by the selector value $n$. The set of ordered pairs $\left\langle \lambda+\Delta\lambda,n+\Delta n\right\rangle $ may serve as coordinates near $\mathscr{M}.$ In these coordinates,
$\mathscr{M}$ has coordinates $\left\langle \lambda,n\right\rangle $.
Define the vectors $\mathbf{u}=\frac{\partial}{\partial\lambda}$
and $\mathbf{n}=\frac{\partial}{\partial n}$. The separation between
the fiducial point $\left\langle \lambda,n\right\rangle $ and the
point with the same $\lambda$ value on the geodesic designated by
$n+\Delta n$ is $\vec{n}=\Delta n\mathbf{n}$. The separation $\vec{n}$
is then parallel transported along the geodesic $n$ by $\Delta\lambda,$
so that the tail of its image is at $\mathscr{N}$ and the tip is
at $\mathscr{B}$. A similar image is produced by parallel transport
of $\vec{n}$ along the geodesic $n$ by $-\Delta\lambda,$ with tail
and tip designated $\mathscr{L}$ and $\mathscr{A}$ respectively.
Beginning at the point $\mathscr{Q}$ which has coordinates $\left\langle \lambda,n+\Delta n\right\rangle$
the point $\mathscr{R}$ is determined by a parameter change $\Delta\lambda$
along the geodesic $n+\Delta n.$ The point $\mathscr{P}$ is determined
by a change $-\Delta\lambda$ along the same $n+\Delta n$ geodesic.
The points $\mathscr{A}$ and $\mathscr{B}$ are determined by parallel
transporting $\vec{n}=\Delta n\mathbf{n}$ along the geodesic $n$
by $-\Delta\lambda$ and $\Delta\lambda$ respectively. The vectors
$\mathscr{B}\mathscr{R}$ and $\mathscr{A}\mathscr{P}$ are then parallel
transported (along unspecified routs) to $\mathscr{Q}$ were they
are summed to produce 
$$
\delta_{2}=\mathscr{B}\mathscr{R}+\mathscr{A}\mathscr{P}=\left(\Delta\lambda\right)^{2}\Delta n\left(\nabla_{\mathbf{u}}\nabla_{\mathbf{u}}\mathbf{n}\right),
$$
where $\nabla_{\mathbf{u}}\nabla_{\mathbf{u}}\mathbf{n}$ is defined
to be the relative-acceleration vector.
On page 271 we find:

[Examine Box 11.4] Thereby arrive at the remarkable equation (11.6)
$$
\nabla_{\mathbf{u}}\nabla_{\mathbf{u}}\mathbf{n}+\left[\nabla_{\mathbf{n}}\nabla_{\mathbf{u}}\right]\mathbf{u}=0.
$$
This equation is remarkable, because at first sight it seems crazy.
  The term $\left[\nabla_{\mathbf{n}}\nabla_{\mathbf{u}}\right]\mathbf{u}$
  involves second derivatives of $\mathbf{u}$ and first derivatives
  of $\nabla_{\mathbf{n}}:$
$$
\left[\nabla_{\mathbf{n}}\nabla_{\mathbf{u}}\right]\mathbf{u}=\nabla_{\mathbf{n}}\nabla_{\mathbf{u}}\mathbf{u}-\nabla_{\mathbf{u}}\nabla_{\mathbf{n}}\mathbf{u}.
$$
It thus must depend on how $\mathbf{u}$ and $\mathbf{n}$vary from
  point to point. But the relative acceleration it produces, $\nabla_{\mathbf{u}}\nabla_{\mathbf{u}}\mathbf{n},$
  is known to depend only on the values of $\mathbf{u}$ and $\mathbf{n}$
  at the fiducial point, not on how $\mathbf{u}$ and $\mathbf{n}$
  vary (see box 11.2 F).

I do not understand what this means. The relative acceleration is
derived by multiple parallel transport steps, and taking differences
of vectors at different locations. That is, its establishment involves
more than simply the values of $\mathbf{u}$ and $\mathbf{n}$ at
one point. Therefore, if we interpret "the fiducial point" to
mean, $\mathscr{Q}$ alone, the statement doesn't make sense to me,
at all.
Even if we allow "the fiducial point" to mean any arbitrary point
on the fiducial geodesic, the derivation still involves points not
on that geodesic, and not determined solely by parallel transport
along the fiducial geodesic.
What does it mean to say "$\nabla_{\mathbf{u}}\nabla_{\mathbf{u}}\mathbf{n},$
is known to depend only on the values of $\mathbf{u}$ and $\mathbf{n}$
at the fiducial point, not on how $\mathbf{u}$ and $\mathbf{n}$
vary"?
PS: I believe the essential conclusion is that relative acceleration can be expressed as the operation of a multilinear form (tensor field $\mathbf{\text{Riemann}}$) on $\mathbf{u}$ and $\mathbf{n}$.  What I'm not getting is the reasoning leading to that conclusion.  
The situation seems similar to that of the differential of a multivariable mapping being a linear mapping associated with its derivative matrix.

 A: Just means it's linear in u and n.
Edit 
Your relative acceleration vector is just geodesic deviation. It is the physical manifestation of 
the riemann tensor
 R(u, n)u which is a vactor valued 2-form. It tells you the deviation of a geodesic in the neighborhood of n connected by u, at each point p(t) where t is the affine parameter corresponding to the proper time. Iff the rieamann tensor is nonzero inertial geodesics will accelerate with respect to each other (tidal effect). 
So you just want the deviation at a given point, not the deviation of the deviation. That's why the tensor contains no derivatives of u and n. So R(lu, n) u =lR(u, n) u where l is a scalar function. 
A: I believe I finally figured this out.  If we think in terms of a covariantly constant tangent plane moved around the point of evaluation $\mathscr{P}_0$ (I think this is what Cartan called a mobile frame), the second covariant derivative can be thought of as the difference between first covariant derivatives at $\mathscr{P}_\lambda$ and $\mathscr{P}_0$ divided by $\lambda,$ the transport path increment in the limit $\lambda\to{0}$.  In that interpretation, $\mathbf{u}$ and $\mathbf{n}$ are the coordinate induced basis vectors of the tangent plane, which are non-constant fields.  In Box 11.2 everything is done with respect to a single, fixed tangent plane.  The basis vectors are parallel transported individually, rather than as a coordinate frame.
It seems worth observing that we only take the difference between $\mathbf{n}_\parallel$ and $\mathbf{n}_{\pm\Delta\lambda}$  We never explicitly take the covariant derivative of $\mathbf{u}$ since it is a universal covariant constant.
