When constructing a general Lorentz boost using an $x$-axis boost, what is the second rotation in relation ro the first rotation?

The solution is a lot simpler than it seems. The result from the first link that you provide is true even in general, and the suggestion in the second link that the rotations are unrelated is incorrect. In general, the relationship between the two rotations in $Lambda=R_2Lambda_xR_1$ is that $R_1$ and $R_2$ are inverses (which is to say, transposes) of one another.

Now, let's justify that answer.

All of the operations that we encounter in this problem are finite-dimensional (i.e. 4-dimensional) and linear, which means that they can be represented as $4times4$ matrices. Generally, there are two interpretations that can be given to a non-singular square matrix: (1) it is a change of basis from one coordinate system to another, or (2) it is a linear transformation which maps vectors in a linear vector space to other vectors in that same space. In this problem, we are obviously interpreting the rotations as changes of basis and the boost as a linear transformation.

Given two bases $A$ and $B$ for some linear vector space and a linear transformation $T$ on that space, it is well known that $T$'s representations in the two different bases are related by $T_B=U_{Ato B}T_AU_{Bto A}$ (where $U_{Ato B}$ is the change of basis matrix from $A$ to $B$). It is clear that by definition $U_{Ato B}=(U_{Bto A})^{-1}$. For any rotation matrix $R$, $R^{-1}=R^T$; therefore, it is now obvious that a Lorentz boost along any arbitrary axis can be given by $R^TLambda_xR$

This answers the question as posed, but for the sake of concreteness, let's derive the general result for an arbitrary Lorentz boost using this method.

First, we will use the formula provided in this Math StackExchange answer to calculate the form of the rotation matrices.

Since we want to rotate the unit vector $hat n = <n_x,n_y,n_z>$ into the unit vector $hat x=<1,0,0>$, we get $hat n cdot hat x = cos(theta)= n_x$ and $hat n times hat x=<0,n_z,-n_y>$. This gives

$$ [v]_times=left[ begin {array}{ccc} 0&n_y&n_z -n_y&0&0 -n_z&0&0end {array} right] $$

Therefore, from the Math StackExchange answer, we get

$$R = I + [v]_{times} + [v]_{times}^2frac{1}{1+cos(theta)}$$

and we derive that the $4times4$ rotation matrix is

$$ R(hat n,hat x) = left[ begin{array}{cccc} 1&0&0&0 0&n_x&n_y&n_z 0&-n_y&1-frac{n_y^2}{1+n_x}&-frac{n_y n_z}{1+n_x} 0&-n_z&-frac{n_y n_z}{1+n_x}&1-frac{n_z^2}{1+n_x} end{array}right] $$

As a note, $R^T(hat n,hat x)=R(hat x, hat n)$ as expected.

The matrix for the $x$-axis Lorentz boost is

$$ Lambda(betahat x) = left[begin{array}{cccc} gamma&-gammabeta&0&0 -gammabeta&gamma&0&0 0&0&1&0 0&0&0&1 end{array}right] $$

This leads to the final calculation of the matrix product

$$ Lambda(betahat n)=R^T(hat n,hat x)Lambda(betahat x)R(hat n,hat x) $$

After some tedious algebra, the final result is

$$ Lambda(betahat n) = left[begin{array}{cccc} gamma&-gammabeta n_x&-gammabeta n_y&-gammabeta n_z -gammabeta n_x&1+(gamma - 1) n_x^2&(gamma - 1)n_xn_y&(gamma - 1)n_xn_z -gammabeta n_y&(gamma - 1)n_xn_y&1+(gamma - 1) n_y^2&(gamma - 1)n_yn_z -gammabeta n_z&(gamma - 1)n_xn_z&(gamma - 1)n_yn_z&1+(gamma - 1) n_z^2 end{array}right] $$

which is (modulo notation) this boost matrix, which is the standard result cited in e.g. Jackson.