I don't see how one would formally relate the two in the above example

Let $$ \hat{\sigma}\cdot{\underline{n}}= n^1\hat{\sigma}_1 + n^2\hat{\sigma}_2+n^3\hat{\sigma}_3 = \hat{S}(\theta,\phi)\hat{\sigma}_3 $$ namely any combination of Pauli matrices can always be written as the product of one other Pauli matrix times another object in $SU(2)$. The matrix $\hat{S}(\theta,\phi)$ is, in the case at hand, the element $$ \hat{S}(\theta, \phi) = \begin{pmatrix} e^{-i\phi/2}\cos(\theta/2)\quad e^{-i\phi/2}\sin(\theta/2)\\ e^{i\phi/2}\sin(\theta/2)\quad e^{i\phi/2}\cos(\theta/2) \end{pmatrix}. $$ Now let us associate two vectors in $\mathbb{R}^3$ to the above equation, one to the left hand side and one to the right hand side, namely $$ (n^1, n^2, n^3)\mapsto(0,0,n'^3) $$ as the right hand side only possesses the $\sigma_3$. The transformation matrix $\hat{R}(\theta, \phi)$ such that $$ \hat{R}(\theta, \phi)\,(n^1, n^2, n^3) = (0,0,n'^3). $$ As such we have now two objects at our disposal: $$ \hat{S}(\theta, \phi)\in SU(2),\qquad \hat{R}(\theta,\phi)\in SO(3); $$ the map $\rho\colon S\mapsto R$ for each pair $(\theta, \phi)$ is the desired double covering map (one can show that both $S$ and $-S$ correspond to the same $R$).

Because of the above argument multiplying the $\sigma$ by $\hat{S}$ is equivalent to multiplying $\underline{n}$ by $\hat{R}$, that is why the eigenvalues are the same no matter whichever multiplication you perform first.

I don't see how one would formally relate the two in the above example

Let $$ \hat{\sigma}\cdot{\underline{n}}= n^1\hat{\sigma}_1 + n^2\hat{\sigma}_2+n^3\hat{\sigma}_3 = \hat{S}(\theta,\phi)\hat{\sigma}_3 $$ namely any combination of Pauli matrices can always be written as the product of one other Pauli matrix times another object in $SU(2)$. The matrix $\hat{S}(\theta,\phi)$ is, in the case at hand, the element $$ \hat{S}(\theta, \phi) = \begin{pmatrix} e^{-i\phi/2}\cos(\theta/2)\quad e^{-i\phi/2}\sin(\theta/2)\\ e^{i\phi/2}\sin(\theta/2)\quad e^{i\phi/2}\cos(\theta/2) \end{pmatrix}. $$ Now let us associate two vectors in $\mathbb{R}^3$ to the above equation, one to the left hand side and one to the right hand side, namely $$ (n^1, n^2, n^3)\mapsto(0,0,n'^3) $$ as the right hand side only possesses the $\sigma_3$. Let $\hat{R}(\theta, \phi)$ the transformation matrix such that $$ \hat{R}(\theta, \phi)\,(n^1, n^2, n^3) = (0,0,n'^3). $$ As such we have now two objects at our disposal: $$ \hat{S}(\theta, \phi)\in SU(2),\qquad \hat{R}(\theta,\phi)\in SO(3); $$ the map $\rho\colon S\mapsto R$ for each pair $(\theta, \phi)$ is the desired double covering map (one can show that both $S$ and $-S$ correspond to the same $R$).

Because of the above argument multiplying the $\sigma$ by $\hat{S}$ is equivalent to multiplying $\underline{n}$ by $\hat{R}$, that is why the eigenvalues are the same no matter whichever multiplication you perform first.

I don't see how one would formally relate the two in the above example

Let $$ \hat{\sigma}\cdot{\underline{n}}= n^1\hat{\sigma}_1 + n^2\hat{\sigma}_2+n^3\hat{\sigma}_3 = \hat{S}(\theta,\phi)\hat{\sigma}_3 $$ namely any combination of Pauli matrices can always be written as the product of one other Pauli matrix times another object in $SU(2)$. The matrix $\hat{S}(\theta,\phi)$ is, in the case at hand, the element $$ \hat{S}(\theta, \phi) = \begin{pmatrix} e^{-i\phi/2}\cos(\theta/2)\\ e^{-i\phi/2}\sin(\theta/2) \end{pmatrix}. $$ Now let us associate two vectors in $\mathbb{R}^3$ to the above equation, one to the left hand side and one to the right hand side, namely $$ (n^1, n^2, n^3)\mapsto(0,0,n'^3) $$ as the right hand side only possesses the $\sigma_3$. The transformation matrix $\hat{R}(\theta, \phi)$ such that $$ \hat{R}(\theta, \phi)\,(n^1, n^2, n^3) = (0,0,n'^3). $$ As such we have now two objects at our disposal: $$ \hat{S}(\theta, \phi)\in SU(2),\qquad \hat{R}(\theta,\phi)\in SO(3); $$ the map $\rho\colon S\mapsto R$ for each pair $(\theta, \phi)$ is the desired double covering map (one can show that both $S$ and $-S$ correspond to the same $R$).

Because of the above argument multiplying the $\sigma$ by $\hat{S}$ is equivalent to multiplying $\underline{n}$ by $\hat{R}$, that is why the eigenvalues are the same no matter whichever multiplication you perform first.

I don't see how one would formally relate the two in the above example

Let $$ \hat{\sigma}\cdot{\underline{n}}= n^1\hat{\sigma}_1 + n^2\hat{\sigma}_2+n^3\hat{\sigma}_3 = \hat{S}(\theta,\phi)\hat{\sigma}_3 $$ namely any combination of Pauli matrices can always be written as the product of one other Pauli matrix times another object in $SU(2)$. The matrix $\hat{S}(\theta,\phi)$ is, in the case at hand, the element $$ \hat{S}(\theta, \phi) = \begin{pmatrix} e^{-i\phi/2}\cos(\theta/2)\quad e^{-i\phi/2}\sin(\theta/2)\\ e^{i\phi/2}\sin(\theta/2)\quad e^{i\phi/2}\cos(\theta/2) \end{pmatrix}. $$ Now let us associate two vectors in $\mathbb{R}^3$ to the above equation, one to the left hand side and one to the right hand side, namely $$ (n^1, n^2, n^3)\mapsto(0,0,n'^3) $$ as the right hand side only possesses the $\sigma_3$. The transformation matrix $\hat{R}(\theta, \phi)$ such that $$ \hat{R}(\theta, \phi)\,(n^1, n^2, n^3) = (0,0,n'^3). $$ As such we have now two objects at our disposal: $$ \hat{S}(\theta, \phi)\in SU(2),\qquad \hat{R}(\theta,\phi)\in SO(3); $$ the map $\rho\colon S\mapsto R$ for each pair $(\theta, \phi)$ is the desired double covering map (one can show that both $S$ and $-S$ correspond to the same $R$).

Because of the above argument multiplying the $\sigma$ by $\hat{S}$ is equivalent to multiplying $\underline{n}$ by $\hat{R}$, that is why the eigenvalues are the same no matter whichever multiplication you perform first.