= Solution
The <geodesic deviation> equation compares neighboring <geodesics> by <covariant derivatives>. Replacing those derivatives by ordinary derivatives of coordinate components requires a <parallel transport> frame along the reference <geodesic>. There is also an <affine parameter> normalization to specify: the argument of $M$ is $u$, and it becomes $\tau$ only after choosing $u=\tau$.
For the intended local tidal calculation, take the reference <timelike geodesic>
$$
\gamma(\tau)=(u,v,x,y)=(\tau,-\tau/2,0,0),
\qquad
V=\partial_u-\frac12\partial_v.
$$
Here $g(V,V)=-1$, so $\tau$ is <proper time>. Both $F$ and its first <partial derivatives> vanish on $\gamma$. All <Christoffel symbols> therefore vanish along this curve, and its <coordinate basis> is <parallel transport> along it. With the <Riemann curvature tensor> from part (b),
$$
R^i{}_{uju}=-M_{ij},\qquad
R^u{}_{bcd}=0,\qquad
R^v{}_{uju}=0\quad\text{on }\gamma.
$$
The <Jacobi field> equation thus becomes, in this adapted frame,
$$
\boxed{
\ddot\sigma^u=0,\qquad
\ddot\sigma^v=0,\qquad
\ddot\sigma^i=M_{ij}(\tau)\sigma^j.
}
$$
A purely transverse initial <Jacobi field>, with $\sigma^u=\sigma^v=\dot\sigma^u=\dot\sigma^v=0$ initially, remains transverse. This <transverse tidal equation of a Brinkmann plane wave> describes the relative <tidal acceleration> of the freely falling bodies.
In vacuum write the <symmetric matrix> as
$$
M(u)=
\begin{pmatrix}
A(u)&B(u)\\ B(u)&-A(u)
\end{pmatrix}.
$$
Its instantaneous <eigenvalues> are $\pm\sqrt{A^2+B^2}$. The <tidal acceleration> stretches along one principal axis and squeezes along the perpendicular axis, with the axes allowed to vary with $u$. This describes accelerations; it does not imply monotonically growing and shrinking separations for arbitrary initial velocities. For the simple constant profile $A=\omega^2$, $B=0$,
$$
\sigma^x=C\cosh(\omega\tau)+D\sinh(\omega\tau),
\qquad
\sigma^y=P\cos(\omega\tau)+Q\sin(\omega\tau).
$$
Thus the <geodesic deviation> can be defocusing in one direction and oscillatory in the other. For a varying profile, diagonalizing at each instant does not decouple the <ordinary differential equations> without also including derivatives of the rotating <eigenvectors>.
\b[Qualification of the coordinate claim.] The displayed equation for $\sigma^v$ is not true for arbitrary coordinate displacements along arbitrary reference <geodesics>. For synchronized neighboring curves with $u=\tau$ and $\sigma^u=0$, the exact <geodesic equations> are
$$
\ddot{\mathbf x}=M\mathbf x,\qquad
\ddot v=-\frac12\mathbf x^{T}M'\mathbf x
-2\dot{\mathbf x}^{T}M\mathbf x.
$$
Their variation gives
$$
\ddot{\boldsymbol\sigma}_{\perp}=M\boldsymbol\sigma_{\perp},\qquad
\ddot\sigma^v
=-\mathbf x^{T}M'\boldsymbol\sigma_{\perp}
-2\dot{\mathbf x}^{T}M\boldsymbol\sigma_{\perp}
-2\mathbf x^{T}M\dot{\boldsymbol\sigma}_{\perp}.
$$
The second expression need not vanish. A concrete vacuum counterexample is $M=\operatorname{diag}(1,-1)$ and the one-parameter family of <timelike geodesics>
$$
u=\tau,\qquad
x=Ae^\tau,\qquad
y=0,\qquad
v=-\frac{\tau}{2}-\frac{A^2}{2}e^{2\tau}.
$$
Each has $g(V,V)=-1$. Differentiating with respect to $A$ produces a <Jacobi field>
$$
\sigma^u=0,\qquad
\sigma^x=e^\tau,\qquad
\sigma^y=0,\qquad
\sigma^v=-Ae^{2\tau},
\qquad
\ddot\sigma^v=-4Ae^{2\tau}.
$$
This even has $g(V,\sigma)=0$, so it is a physical orthogonal separation. The <covariant derivative> form of <geodesic deviation> remains correct; the unrestricted assertion about ordinary coordinate derivatives needs the adapted reference and frame used above. If instead $\dot u=c$ is a constant <affine parameter> normalization, the transverse tidal coefficient is $c^2M(u)$.
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