Transpose of a bilinear in Einstein notation In Einstein notation we can take generic 1-vectors $x, y$ and (1,1) tensor $M$.
As we know $x_{\mu}$ represents $x^{T}$, i.e. row vector (a co-vector), while $x^{\mu}$ is a column vector.
So we can take the bilinear:
$$x^{T}My=x_{\mu}{M^{\mu}}_{\nu}y^{\nu}\tag{1}$$
Since this bilinear is a scalar, it is true that:
$$x^{T}My=(x^{T}My)^{T}=y^{T}M^{T}x\tag{2}$$
So far so good.
The problem comes when I try to represent the last expression using the index notation, indeed I can "derive naively" the last equation in index notation, simply using $AB=A^{\rho}B_{\rho}=A_{\rho}B^{\rho}$ and the definition of transpose matrix: $${M^{\mu}}_{\nu}={(M^T)_{\nu}}^{\mu}\tag{3}$$
So we have:
$$x^{T}My=x_{\mu}{M^{\mu}}_{\nu}y^{\nu}=y^{\nu}{M^{\mu}}_{\nu}x_{\mu}=y^{\nu}{(M^T)_{\nu}}^{\mu}x_{\mu}=y_{\nu}{(M^T)^{\nu}}_{\mu}x^{\mu}=y^{T}M^{T}x\tag{4}$$
But this is wrong, because for example for a Lorentz Transformation (but it is true in general) we have:
$$\Lambda^T\eta\Lambda ~=~ \eta \quad\Rightarrow\quad 
(\Lambda^T)_{\rho}{}^{\mu}~\eta_{\mu \nu}~ \Lambda^{\nu}{}_{\sigma} ~=~ \eta_{\rho \sigma} \tag{5}$$
This means that in the expression $y^{T}M^{T}x$ it is incorrect to write ${(M^T)^{\nu}}_{\mu}$ because $(M^T)$ should be represented as ${(M^T)_{\mu}}^{\nu}$, as we did in (5), so my question are:
A) Which is the correct way to represent $y^{T}M^{T}x$ ?
I think it should be:
$y^{T}M^{T}x=y_{\nu}{(M^T)_{\mu}}^{\nu}x^{\mu} \tag{6}$
But I am not sure, and if it is then:
B) What is the error in (4)? i.e. Which is the correct derivation to get the expression (2) in Einstein Notation?
 A: I'm not sure I completely understand your question, but here are some thoughts that I hope you will find useful.
You have to be a little careful when translating between index notation and ordinary matrix notation.

As we know $x_\mu$ represents $x^T$, i.e. row vector (a co-vector), while $x^\mu$ is a column vector.

It may sometimes be helpful to think of $x_\mu$ as analogous to a row-vector and $x^\mu$ as analogous to a column-vector (or vice versa), but the analogy should not be taken too far. To illustrate this, consider the scalar $x^\mu \eta_{\mu \nu} x^\nu$. To represent this expression using matrix-vector notation, you would represent $\eta$ as a 4x4 matrix, and the two $x$'s as vectors. However, you would then have to represent either $x^\mu$ or $x^\nu$ as a row vector for the expression to make sense; if you blindly take both of them to be column vectors and $\eta_{\mu \nu}$ to be a matrix, you end up with a nonsensical expression.
The Einstein notation is more general than the matrix- and vector notation you may be used to, and thus contains different objects which cannot readily be distinguished in matrix-vector notation if you're not careful.
I think an example may be helpful.
Consider a rank 2 tensor $M$. Being a rank 2 tensor, it has two indices which we can raise and lower using the Minkowski tensor,
$$M_{\mu \nu} = \eta_{\mu \alpha} {M^\alpha}_\nu 
=\eta_{\mu \alpha}\eta_{\nu \beta } {M^\alpha}^\beta = \eta_{\nu \alpha} {M_\mu}^\alpha,$$
etc.
We want to represent this tensor as a 4x4 matrix, but there are many ways we could do this.
For instance we could choose to look at the elements ${M^\mu}_\nu$, where the first index is raised and the second is lowered; then let the first index $\mu$ label rows and the second index $\nu$ columns of our matrix. Thus we obtain the matrix
\begin{equation}
\label{eq:1}
M \longrightarrow \begin{bmatrix}
{M^{0}}_{0} & {M^{0}}_{1} & {M^{0}}_{2} & {M^{0}}_{3} \\
{M^{1}}_{0} & {M^{1}}_{1} & {M^{1}}_{2} & {M^{1}}_{3} \\
{M^{2}}_{0} & {M^{2}}_{1} & {M^{2}}_{2} & {M^{2}}_{3} \\
{M^{3}}_{0} & {M^{3}}_{1} & {M^{3}}_{2} & {M^{3}}_{3}
\end{bmatrix}.
\end{equation}
However, we could also have let the second index label the rows, or chosen to work with $M_{\mu \nu}$ or ${M_\mu}^\nu$ instead. If we had done so, it would have resulted in a different matrix representing the same tensor $M$. Thus, the above relation expresses just one way to relate a rank-2 tensor to a matrix.
The transpose of the 4x4 matrix above is
$$\begin{bmatrix}
{M^{0}}_{0} & {M^{1}}_{0} & {M^{2}}_{0}  & {M^{3}}_{0} \\
{M^{0}}_{1} & {M^{1}}_{1} & {M^{2}}_{1} & {M^{3}}_{1} \\ 
{M^{0}}_{2} & {M^{1}}_{2} & {M^{2}}_{2} & {M^{3}}_{2} \\
{M^{0}}_{3} & {M^{1}}_{3} & {M^{2}}_{3} & {M^{3}}_{3} 
\end{bmatrix}.$$
Under our map defined before, we see that this matrix is the image of the rank-2 tensor $A$ with elements
${A^\mu}_{\nu} = {M^\nu}_\mu$. We could then in a somewhat reasonable sense say that the tensor $A$ is the "transpose" of $M$. However, this  expression is not Lorentz- or basis-invariant, as the upper and lower indices do not match on either side of the expression, and it also doesn't match the definition you give in equation (3).
To match (3), we would need to say that the tensor with elements
${A_\mu}^\nu = {M^\nu}_\mu$
represents the "transpose" of $M$. However note that to represent this $A$ as a matrix, we need to use a diffferent transformation rule when converting it to a matrix, in order for the obtained matrix to actually be the transpose of the matrix obtained for $M$.
In your equation (4) all of the expressions using indices are equivalent, so they are all "correct" - however, if you wish to transform the index-expression to a matrix expression, you need to choose a transformation rule - and depending on the rule you choose, you may obtain different matrix expressions. Of course, the scalar that you obtain in the end will always be the same.
