What’s the relation between local diagonalisation of Riemannian metrics and the eigenprojections for the representation matrix functions?

From my understanding, the diagonalisation of $g_p$ would then be equivalent to finding smooth eigenprojections for the representation matrix $T(x)$.

Your understanding is false already in the case $n=2$. I suggest you start by reading a proof of existence of isothermal coordinates in the 2d case. The difficult part to find a local diffeomorphism whose derivative at every $x$ sends eigenlines of $g_x$ in $T_xM$ to coordinate lines in the 3d space. In the 2d case this is relatively easy (generically) since the eigenlines of $g_x$ determine two transversal 1-dimensional foliations in the plane and then you can map these foliations to the foliations by vertical and horizontal lines in the plane. This argument though breaks down already in the 2d case at a point where $g_x$ is already a scalar multiple of the Euclidean metric. The 3d case is even harder.

Edit. As for your example from Kato's book: The matrices $T(x)$ satisfy $T(0)=0$, which means that for $x=0$ the splitting to $R^2$ into a pair of 1-dimensional eigenspaces is undefined. (The eigenvalues are equal.) Therefore, you should not expect the the eigenprojections to have a continuous extension to $x=0$. (The "genericity" assumption I made earlier means that the eigenvalues at every point are distinct.) The bottom line is that "diagonalization theorems" for metrics have very little to do with eigenprojections.

As Moishe pointed out, you need to be careful to make the distinction between orthogonal/isothermal coordinates and orthogonal frames - the former is much stronger, while the latter always exist locally (by Gram-Schmidt). However, I think the crux of your question (the non-equivalence of a diagonalizing frame and frames aligned to the eigenspaces) still stands.

I believe the key issue is the distinction between the usual matrix diagonalization by the similarity relation $g = PDP^{-1}$, and the diagonalization by the congruence relation $g = PDP^T$, which is what's going on here: we're diagonalizing a symmetric bilinear form, without any requirement that the diagonalizing basis is orthonormal. (Indeed, since we're attempting to diagonalize the metric, we don't have any natural "background" inner product.)

For similarity diagonalization, the diagonalizing basis $P$ is generically unique, and thus the regularity questions that Kato's book investigates are very natural. Congruence diagonalization of symmetric bilinear forms by orthogonal transformations is equivalent to similarity diagonalization (since $P^T = P^{-1}$ for orthogonal $P$), but if you're not making this restriction on $P$ then the situation is highly non-unique: if $g = PDP^T$ is positive definite, we can write this in the canonical form $g = (P \sqrt D) I (P \sqrt D)^T$, and in fact we have $g = (P \sqrt D O) I (P \sqrt D O)^T$ for any orthogonal $O.$ Thus the fact that $g$ is congruence-diagonalized in a given frame does not imply that this frame is aligned with the eigenprojections of $g$.

For a concrete example, consider the symmetric positive-definite matrix $$g=\left(\begin{array}{cc} 2 & 1\\ 1 & 2 \end{array}\right),$$ which has orthogonal diagonalization $$ g=\left(\begin{array}{cc} \frac{1}{\sqrt{2}} & -\frac{1}{\sqrt{2}}\\ \frac{1}{\sqrt{2}} & \frac{1}{\sqrt{2}} \end{array}\right)\left(\begin{array}{cc} 3 & 0\\ 0 & 1 \end{array}\right)\left(\begin{array}{cc} \frac{1}{\sqrt{2}} & -\frac{1}{\sqrt{2}}\\ \frac{1}{\sqrt{2}} & \frac{1}{\sqrt{2}} \end{array}\right)^{-1}$$

but also many other congruence-diagonalizations, including

$$g=\left(\begin{array}{cc} 1 & 0\\ \frac{1}{2} & 1 \end{array}\right)\left(\begin{array}{cc} 2 & 0\\ 0 & \frac{3}{2} \end{array}\right)\left(\begin{array}{cc} 1 & 0\\ \frac{1}{2} & 1 \end{array}\right)^{T}.$$

Thus it shouldn't be too surprising that we can smoothly diagonalize a parametrized bilinear form in this sense, since we have a lot more room to move. Indeed, we always have existence of local orthonormal frames - you just take the usual Gram-Schmidt construction of an orthonormal basis, apply it to a smooth local frame and check that everything is smooth.