$\newcommand{\ket}[1]{\left| #1 \right>}$$\newcommand{\bk}[2]{\left< #1 | #2 \right>}$Notice that the eigenvectors of the operator $S_z$ spans the whole space, which means that you can write any state as a superposition, (if you prefer as a linear combination) of these states. The situation is akin to the basis vectors of usual 3d Euclidean space. You can choose tree basis vectors there, usually the following is chosen:
$$\vec x = \begin{pmatrix} 1 \\ 0 \\ 0 \end{pmatrix}\quad \vec y = \begin{pmatrix} 0 \\ 1 \\ 0 \end{pmatrix}\quad \vec z = \begin{pmatrix} 0 \\ 0 \\ 1 \end{pmatrix}\quad$$
Then you can write any vector $\vec v$ as a linear combination of these vectors that is
$$\vec v = \begin{pmatrix} a \\ b \\ c \end{pmatrix} \implies \vec v = a \vec x + b \vec y + c \vec z$$
Notice that the choice of basis vectors is not unique. I could have chosen the following basis vectors if I wanted to:
$$\vec x' = \begin{pmatrix} 1 \\ 1 \\ 0 \end{pmatrix}\quad \vec y' = \begin{pmatrix} 0 \\ 1 \\ 0 \end{pmatrix}\quad\vec z' = \begin{pmatrix} 0 \\ 0 \\ 1 \end{pmatrix}\quad$$
for which the vector $\vec v$ can be written as
$$\vec v = a\vec x' + (b-a) \vec y' + c\vec z'$$
The situation with spin 1/2 particle is almost the same. You can choose your basis vectors to be the eigenvectors of $S_z$ operator in which case you write
$$\ket{z+} = \begin{pmatrix} 1 \\ 0 \end{pmatrix} \quad \ket{z-} = \begin{pmatrix} 0 \\ 1 \end{pmatrix} $$
$$\implies \ket{x+}= \frac{1}{\sqrt 2} \Big( \ket{z+}+ \ket{z-} \Big)\quad \ket{x-}= \frac{1}{\sqrt 2} \Big( -\ket{z+} + \ket{z-} \Big)$$
Notice that $\ket{x\pm}$ and $\ket{z\pm}$ are the eigenvectors of $S_x$ and $S_z$ operators respectively, as we wanted.
If you send a particle with spin $+x$ in a SG machine aligned in the $z$ direction, then half the time you get $+z$ and half the time you get $-z$. You can immediately see that by taking the inner product of the state $\ket{x+}$ with the states $\ket{ z \pm}$.
$$\left| \bk{z+}{x+} \right|^2 = \left( \frac{1}{\sqrt 2} \right)^2 = \frac{1}{2} \qquad{\text{and similarly}} \quad \left| \bk{z-}{x+} \right|^2 = \frac{1}{2}$$
You can also choose your basis vectors to be the eigenvectors of $S_x$ operator but everything gets messy for this problem and you don't, in general, want to do that.