Einstein comment about coordinates I am reading Einstein's note about genesis of GR, and I think I did not understand the part below, from the point where he starts talking about time coordinate. In particular, I don't understand this part:
When using non-linear transformations "the immediate metrical significance of the coordinates is lost"

"We start with an empty, field-free space as it appears with respect to an inertial system in accord with the special theory of relativity, as the simplest of all conceivable physical situations. Now if we imagine a non-inertial system introduced in such a way that the new system (described in three-dimensional language) is uniformly accelerated in a (suitably defined) direction with respect to the inertial system; then, with respect to this system, there exists a static parallel gravitational field. In this case, the reference system may be chosen as a rigid one, in which three-dimensional Euclidean metric relations hold. But that time [coordinate–JS], in which the field appears static, is not measured by equally constituted clocks at rest [in that system–JS]. From this special example, one already recognizes that, when one allows non-linear transformations of any sort, the immediate metrical significance of the coordinates is lost. One must introduce such transformations, however, if one wants to justify the equality of gravitational and inertial mass by the foundations of the theory, and if one wants to overcome Mach’s paradox concerning inertial systems."

 A: 
In particular, I don't understand this part: When using non-linear transformations "the immediate metrical significance of the coordinates is lost"

I think what he means is that the squared length element $d\ell^2$ is no longer given by a simple sum of squares of the individual coordinate elements $dx_i^2$.
As a 2D example, suppose $x_i$ are the usual cartesian coordinates $x$ and $y$. Then we have:
$$
d\ell^2 = dx^2 + dy^2\;.
$$
If we make a linear transformation (e.g., rotation) to a different set of coordinates $p$ and $q$ where:
$$
\begin{bmatrix}
           p \\
           q \\
\end{bmatrix}
=
\bar R
\begin{bmatrix}
           x \\
           y \\
\end{bmatrix}
$$
and the matrix $\bar R$ is a rotation matrix then we still have
$$
d\ell^2 = dp^2 + dq^2\;,
$$
since
$$
d\ell^2 = (dx\;dy)\cdot \begin{bmatrix}
           dx \\
           dy \\
\end{bmatrix} 
= \vec (dp\;dq) \bar R^T \cdot \bar R \begin{bmatrix}
           dp \\
           dq \\
\end{bmatrix}
=
(dp\; dq)\cdot \begin{bmatrix}
           dp \\
           dq \\
\end{bmatrix}\;,
$$
because $R^T R = 1$ for rotations.
But if we make a non-linear transformation, for example, if we change to circular coordinates $r$ and $\theta$ where
$$
x = r\cos(\theta)
$$
and
$$
y = r\sin(\theta)
$$
Then we no longer have the squared length as a simple sum of squares:
$$
d\ell^2 \neq dr^2 + d\theta^2\;.
$$
Rather, we have to account for the non-linear transformation with a pre-factor of $r^2$ on one of the terms:
$$
d\ell^2 = dr^2 + r^2 d\theta^2\;.
$$
Another way to say it is that $\theta$ is no longer a "length" or we could say it no longer has "immediate metrical significance." But rather we now need some other function of the other coordinates to create a proper length and give it metrical significance.
