How to prove the covariant derivative cannot be written as an eigendecomposition of the partial derivative?
Let there be given a manifold $(M,\nabla)$ equipped with a (not necessarily torsionfree) tangent bundle connection $\nabla$.
I got the (possibly faulty) impression from reading the first lines in OP's question formulation (v18) that OP is asking:
Is it possible that the local coordinate expression for the covariant derivative of a co-vector/one-form $\lambda =\lambda_a \mathrm{d}x^a$ (in some local coordinate system $x^a$) could be on the form $$\tag{1} (\nabla_c \lambda)_a~=~\frac{\partial y^d}{\partial x^a} \frac{\partial }{\partial x^c}\left(\frac{\partial x^b}{\partial y^d} \lambda_b \right) $$ for some (invertible, smooth, locally defined) functions $y^d=y^d(x)$?
This is equivalent to asking:
Is it possible that the Christoffel symbols$^1$ (in some local coordinate system $x^a$) could be of the form $$\tag{2} \Gamma^{(x)b}_{ca}~=~\frac{\partial^2 y^d}{\partial x^c\partial x^a} \frac{\partial x^b}{\partial y^d} $$ for some (invertible smooth, smooth, locally defined) functions $y^d=y^d(x)$?
Recall that the Christoffel symbol $\Gamma^{\lambda}_{\mu\nu}$ does not transform as a tensor under a local coordinate transformation $x^a \to y^b=y^b(x)$, but rather with an inhomogeneous term, which is built from the second derivative of the coordinate transformation
$$\tag{3}\frac{\partial y^f}{\partial x^c} \Gamma^{(x)c}_{ab} ~=~\frac{\partial y^d}{\partial x^a}\, \frac{\partial y^e}{\partial x^b}\, \Gamma^{(y)f}_{de} +\frac{\partial^2 y^f}{\partial x^a \partial x^b}. $$
Therefore formulation (2) is equivalent of asking:
Does there exists a local coordinate system $y^d$ such the Christoffel symbols vanish $$\tag{4} \Gamma^{(y)c}_{ab}~=~0? $$
This, in turn, is equivalent to asking
Is $(M,\nabla)$ flat?
This is equivalent to asking
Does there exist a local coordinate system $y^d$ such that the metric tensor $$\tag{5} g^{(y)}_{ab}~=~\eta_{ab} $$ is constant?
This is equivalent to asking
In an arbitrary coordinate system $x^a$, is the metric tensor of the form $$\tag{6} g^{(x)}_{ab}~=~\frac{\partial y^c}{\partial x^a}\eta_{cd} \frac{\partial y^d}{\partial x^b} $$ for some (invertible smooth, smooth, locally defined) functions $y^d=y^d(x)$?
Rereading OP's question formulation, condition (6) seems to be one of OP's assumptions. If we are granted assumption (6), then OP's initial claim (1) holds.
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$^1$ It is convenient to call $\Gamma^c_{ab}$ Christoffel symbols even if the tangent-space connection $\nabla$ is not torsionfree.
Equation (13) expresses the metric on an embedded hypersurface given by the relations $y^k = y^k(x^a)$. However, the equation for the inverse metric (4-th equation) is in general not correct.
Take for example a hypersurface defined by: $y^1 = x^1$, $y^2 = x^2$, $ y^3 = x^2$. In our case, the partial derivative of $x^2$ with respect to $y^2$ or $y^3$ cannot be defined. For example, in order to differentiate with respect to $y^2$ we must vary $y^2$ while keeping $y^3$ constant. This is impossible to do on the hypersurface.
Update: This is a detailed answer to linuxfreebird comment.
Although I did not receive my education from Rindler's book, the material of the subject of this question appears in many pedagogical introductions of general relativity, in which an $N$ dimensional space time $X$ is isometrically embedded in a larger flat $M$-dimensional flat space $Y$:
$$\begin{matrix} X \rightarrow Y \\ y^k = y^k(x^1, ...., x^N), k = 1, ..., M \end{matrix}$$
The embedding is given as a set of equations defining $X$ as a hypersurface into $Y$. The functions need not be invertible as we want them to describe embeddings of spaces of different dimensions.
Equation (4) is just the relation expressing that the embedding is isometric. This can be seen more explicitly if we write this equation in the form:
$$ g_{a b} =\eta_{kl} \frac{\partial y^k }{\partial x^a} \frac{\partial y^l}{\partial x^b}$$
Where $\eta$ is the flat metric on the manifold $Y$. Now, this equation looks as a metric transformation, i.e., it preserves distances.
This approach is called extrinsic geometry. It can be used to define the various geometrical objects of the space-time $X$ (such as the Levi-Civita connection and the Riemann curvature) in terms of the embedding equations. This approach is correct, because for any space-time $X$ there exists such an embedding for some high enough $M$ (This is called Nash embedding theorem), but it is seldom used in general relativity real work. Most work in general relativity uses intrinsic geometry without resorting to any embedding equations.
The key point is to remember that the $y$s are functions of the $x$s and every time a partial derivative of $x$ with respect to $y$ is used, means that these equations were inverted, but this cannot be done in general, except in the particular case where these equations are invertible.
Now, in the case when the equations are invertible (it necessarily implies that $ M=N$ ), everything is correct, but this case refers only to space-times that can be isometrically embedded into a flat space-time of the same dimension. These space-times are necessarily flat, thus not general enough to cover interesting cases in general relativity.
May be you can profit from this analysis if you continue your exercise and prove that the Riemann tensor computed from your form of the Levi-Civita connection is globally identically zero.
Special thanks to Qmechanic and David Bar Moshe for providing their answers for this post. Combining Qmechanic's and David Bar Moshe's insights, I finally figured out the confusing element in the problem.
The covariant derivative can be written in terms of an eigendecomposition of the partial derivative, the Christoffel symbols can be written in terms of non-invertible extensive variables, and the Riemman tensor is not necessarily equal to zero. Here is my answer:
I have written the covariant derivative \begin{align} \lambda_{a;c} = \dfrac{\partial y^k}{\partial x^a} \dfrac{\partial }{\partial x^c} \left( \dfrac{\partial y_k}{\partial x^p} g^{pb} \lambda_{b} \right) \end{align} in terms of its eigendecomposition and indices "a" and "c" and again \begin{align} \lambda_{b;d} = \dfrac{\partial y^j}{\partial x^b} \dfrac{\partial }{\partial x^d} \left( \dfrac{\partial y_j}{\partial x^q} g^{qn} \lambda_{n} \right) \end{align} with different indices "b" and "d". Combining the covariant derivatives forms the following: \begin{align} \lambda_{a;cd} =\\ \dfrac{\partial y^k}{\partial x^a} \dfrac{\partial }{\partial x^c} \left( \Lambda^{\ \ j}_{k} \dfrac{\partial }{\partial x^d} \left( \dfrac{\partial y_j}{\partial x^q} g^{qn} \lambda_{n} \right) \right) \end{align} where \begin{align} \Delta^{\ \ j}_{k} = \dfrac{\partial y_k}{\partial x^p} g^{pb} \dfrac{\partial y^j}{\partial x^b} \neq \delta_{k}^{\ \ j} \end{align} because \begin{align} g^{pb} \neq \dfrac{\partial x^p}{\partial y^n} \dfrac{\partial x^b}{\partial y_n} \end{align}
is ill-defined.
The Riemann tensor $R_{acd}^{ \ \ \ \ m} $ is created by computing the commutator of the covariant derviatives as follows:
\begin{align} \lambda_{a;[c,d]} = \left( \dfrac{\partial y^k}{\partial x^a} \dfrac{\partial \Delta^{\ \ j}_{k} }{\partial x^c} \dfrac{\partial }{\partial x^d} \left( \dfrac{\partial y_j}{\partial x^p} g^{pm} \right) \right)_{[c,d]} = R_{acd}^{ \ \ \ \ m} \lambda_m . \end{align}
If the number of dimensions of $y$ equals the number of dimensions in $x$, then $\Delta^{\ \ j}_{k} = \delta^{\ \ j}_{k}$, thus $R_{acd}^{ \ \ \ \ m} = 0$.