I'll consider the special case of symmetric tridiagonal matrices with zero diagonal for this answer.
I prefer calling the even-order tridiagonal ones Golub-Kahan matrices. These matrices turn up in deriving the modification of the QR algorithm for computing the singular value decomposition (SVD). More precisely, given an $n\times n$ bidiagonal matrix like ($n=4$)
$$\mathbf B=\begin{pmatrix}d_1&e_1&&\\&d_2&e_2&\\&&d_3&e_3\\&&&d_4\end{pmatrix}$$
the $2n\times 2n$ block matrix $\mathbf K=\left(\begin{array}{c|c}\mathbf 0&\mathbf B^\top \\\hline \mathbf B&\mathbf 0\end{array}\right)$ is similar to the Golub-Kahan tridiagonal
$$\mathbf P\mathbf K\mathbf P^\top=\begin{pmatrix}& d_1 & & & & & & \\d_1 & & e_1 & & & & & \\& e_1 & & d_2 & & & & \\& & d_2 & & e_2 & & & \\& & & e_2 & & d_3 & & \\& & & & d_3 & & e_3 & \\& & & & & e_3 & & d_4 \\& & & & & & d_4 & \end{pmatrix}$$
where $\mathbf P$ is a permutation matrix. This similarity transformation is referred to as the "perfect shuffle".
The importance of this is that the eigenvalues of the Golub-Kahan matrices always come in $\pm$ pairs; more precisely, if $\mathbf B$ has the singular values $\sigma_1,\sigma_2,\dots,\sigma_n$, then the eigenvalues of the Golub-Kahan tridiagonal are $\pm\sigma_1,\pm\sigma_2,\dots,\pm\sigma_n$.
Odd-order zero-diagonal tridiagonals can be treated similarly, as they have a zero eigenvalue in addition to the $\pm$ pairs of eigenvalues. The treatment given above for Golub-Kahan tridiagonals becomes applicable after deflating out the zero eigenvalue; one can do this by applying the QR decomposition $\mathbf T=\mathbf Q\mathbf R$ and forming the product $\mathbf R\mathbf Q$ and deleting the last row and last column, thus forming a Golub-Kahan tridiagonal.
See Ward and Gray's paper (along with the associated FORTRAN code) and this beautiful survey by David Watkins.
I am not sure whether I understood the question correctly. First of all any $2\times2$ Hermitian matrix $A$ can be written as a real linear combination of the Pauli matrices and they form the basis. To see this, notice that the inner product is given by the trace, and the coefficient $x_s$ of $\sigma_s$ will be $\mathrm{Tr}(A\cdot\sigma_s)$ where $s=0,x,y,z$.
After this you can simply use the fact that for two vector spaces $V_1,~V_2$ with basis $\{e_i\}$ and $\{f_j\}$ respectively, the tensor product $V_1\otimes V_2$ is a vector space with basis elements are given by $\{e_i\otimes f_j\}$.
Best Answer
To prove the proposed equality I will be using the eigenvalue decomposition of the Pauli matrices, which are
where $|0\rangle=\begin{pmatrix}1 \\ 0\end{pmatrix}$ and $|1\rangle=\begin{pmatrix}0 \\ 1\end{pmatrix}$. Now developing the first part of the equality proposed by using the above relationships:
\begin{equation} \begin{split} \frac{1}{4}\sum_{j=0}^3\sigma_jA\sigma_j &=\frac{1}{4}(\sigma_0A\sigma_0+\sigma_1A\sigma_1+\sigma_2A\sigma_2+\sigma_3A\sigma_3) \\ & = \frac{1}{4}[(|0\rangle\langle0|+|1\rangle\langle1|)A|(0\rangle\langle0|+|1\rangle\langle1|)+(|0\rangle\langle1|+|1\rangle\langle0|)A(|0\rangle\langle1|+|1\rangle\langle0|)\\\ & +i^2(|1\rangle\langle0|-|0\rangle\langle1|)A(|1\rangle\langle0|-|0\rangle\langle1|)+(|0\rangle\langle0|-|1\rangle\langle1|)A(|0\rangle\langle0|-|1\rangle\langle1|)]\\ & =\frac{1}{4}[|0\rangle\langle0|A|0\rangle\langle0|+|0\rangle\langle0|A|1\rangle\langle1|+|1\rangle\langle1|A|0\rangle\langle0|+|1\rangle\langle1|A|1\rangle\langle1| \\ & + |0\rangle\langle1|A|0\rangle\langle1|+|0\rangle\langle1|A|1\rangle\langle0|+|1\rangle\langle0|A|0\rangle\langle1|+|1\rangle\langle0|A|1\rangle\langle0| \\ & - (|1\rangle\langle0|A|1\rangle\langle0|+|1\rangle\langle0|A|0\rangle\langle1|+|0\rangle\langle1|A|1\rangle\langle0|+|0\rangle\langle1|A|0\rangle\langle1|) \\ & +|0\rangle\langle0|A|0\rangle\langle0|-|0\rangle\langle0|A|1\rangle\langle1|-|1\rangle\langle1|A|0\rangle\langle0|+|1\rangle\langle1|A|1\rangle\langle1|] \\ & =\frac{1}{4}[2|0\rangle\langle0|A|0\rangle\langle0|+2|1\rangle\langle1|A|1\rangle\langle1|+2|0\rangle\langle1|A|1\rangle\langle0|+2|1\rangle\langle0|A|0\rangle\langle1|] \\ & = \frac{1}{2}[|0\rangle\langle0|A|0\rangle\langle0|+|1\rangle\langle1|A|1\rangle\langle1|+|0\rangle\langle1|A|1\rangle\langle0|+|1\rangle\langle0|A|0\rangle\langle1|] . \end{split} \end{equation}
At this point the effect of the multiplication of those matrices by matrix $A=\begin{pmatrix}a & b \\ c & d\end{pmatrix}$ has to be analyzed:
And so using such relationships and the fact that $tr(A)=a+d$, we continue the derivation started above from the last step
\begin{equation} \begin{split} \frac{1}{4}\sum_{j=0}^3\sigma_jA\sigma_j &=\frac{1}{2}[|0\rangle\langle0|A|0\rangle\langle0|+|1\rangle\langle1|A|1\rangle\langle1|+|0\rangle\langle1|A|1\rangle\langle0|+|1\rangle\langle0|A|0\rangle\langle1|] \\ & = \frac{1}{2}\left[\begin{pmatrix}a & 0 \\ 0 & 0\end{pmatrix} + \begin{pmatrix}0 & 0 \\ 0 & d\end{pmatrix} + \begin{pmatrix}d & 0 \\ 0 & 0\end{pmatrix} + \begin{pmatrix}0 & 0 \\ 0 & a\end{pmatrix} \right]=\frac{1}{2}\begin{pmatrix}a+d & 0 \\ 0 & a+d\end{pmatrix} \\ & = \frac{a+d}{2}\begin{pmatrix}1 & 0 \\ 0 & 1\end{pmatrix}=\frac{tr(A)}{2}I. \end{split} \end{equation}
Note that in the derivation of the equality, the restriction that $A$ must be positive definite has not been used, so the equality holds for all $2\times 2$ matrices.