Intuition for Spin operator in arbitrary direction I understand why the Spin operators in $x$, $y$ and $z$ direction are given by : $\begin{align*}
S_x = \begin{pmatrix}
0 &\hbar/2\\
\hbar/2 & 0
\end{pmatrix}
S_y = \begin{pmatrix}
0 & -i\hbar/2\\
i\hbar/2 & 0
\end{pmatrix}
S_z = \begin{pmatrix}
\hbar/2 & 0\\
0 & -\hbar/2
\end{pmatrix}
\end{align*}$
But why is the spin operator along an arbitrary direction $\vec{n}$ given by : $S_{\vec{n}} = n_x \cdot{S_x} + n_y \cdot{S_y} +n_z \cdot{S_z}$ ?
I can see that it works along the $x$, $y$ and $z$ axis, and that is look like a scalar product between $\vec{n}$ and $ \textbf{S} = (S_x,S_y,S_z)$. I don't need a rigorous proof, a more physical explanation would be ok. I saw this post related, but no satisfactory answer.
EDIT:
Little precision, what is not clear for me is why I can do stuff with $\textbf{S}$ like if it was a vector. Also I would not be satisfied if you just say "it transforms like a vector". It is also not really clear what it would mean to take a scalar product with $\textbf{S}$.
 A: I'm not sure what reasoning took you to accept that,
$$S_{x}=\boldsymbol{e}_{x}\cdot\boldsymbol{S}$$
is OK, but let's take it from there. There's nothing special about “$x$”, as opposed to “$y$”, or “$z$”. You could have said $\boldsymbol{n}$, instead of $\boldsymbol{e}_{x}$, and you would have,
$$S_{\boldsymbol{n}}=\boldsymbol{n}\cdot\boldsymbol{S}$$
IOW, $x$ is a dummy parameter there.
As a simpler example that hopefully will clarify the question (Pauli matrices carry "internal space" representations), consider this:
Suppose you have a physical law that tells you that a certain scalar $\sigma$ that depends on a given direction, x –that's relevant to your problem– is:
$$\sigma=x$$
This is unsatisfactory as a physical law, because it does not have any definite transformation law under rotations. $\sigma$ is a scalar, but $x$ is not. If you want to make it right, you have to upgrade it to a physical law but referring it to an arbitrary direction and make the transformation law transparent:
$$\sigma\left({\boldsymbol{n}}\right)=\boldsymbol{x}\cdot\boldsymbol{n}$$
