What is the vector corresponding to a two-particle product state $|\psi\rangle_1|\psi\rangle_2$? Let's suppose we have two particles described by 
\begin{equation}
|\psi\rangle_1 =  \begin{pmatrix} a \\ b \end{pmatrix}_1 \text{ and } |\psi\rangle_2 =  \begin{pmatrix} \alpha \\ \beta \end{pmatrix}_2
\end{equation}
If the particles don't interact the product state $|\psi\rangle$ of both states describes the whole system.
\begin{equation}
|\psi\rangle=|\psi\rangle_1 |\psi\rangle_2 
\end{equation}
If I know want to write the state like a vector, I get following
\begin{equation}
|\psi\rangle =\begin{pmatrix}
a \\
b \\
\alpha \\
\beta
\end{pmatrix}
\end{equation}
Is this the correct way of doing this? Assuming both one-particle states have length $1$, we get
\begin{equation}
\sqrt{|a|^2+|b|^2+|\alpha|^2 + |\beta|^2}=2
\end{equation}
Obviously the product state should be normed! What is the right way to fix this? Should we define the product state as follows
\begin{equation}
\frac{1}{\sqrt{2}}\begin{pmatrix}
a \\
b \\
\alpha \\
\beta
\end{pmatrix}
\end{equation}
or should we redefine the length of the vector like this
\begin{equation}
\sqrt{|a|^2+|b|^2} \cdot \sqrt{|\alpha|^2 + |\beta|^2} = 1
\end{equation}
Or are both approaches possible? Or are both wrong? Never really found something in the literature because they all stick to the bracket notation.

The tensor product of two 2-entry vectors is not simply a 4-entry vector and does not behave like it, e.g. when you think about things as simple addition of two such objects. For example,
$$\left( \begin{array}{c}  a \\ b  \end{array} \right) \otimes \left( \begin{array}{c}  c \\ d  \end{array} \right) + \left( \begin{array}{c}  a \\ b  \end{array} \right) \otimes \left( \begin{array}{c}  e \\ f  \end{array} \right) = \left( \begin{array}{c}  a \\ b  \end{array} \right) \otimes \left( \begin{array}{c}  c+e \\ d+f  \end{array} \right) \neq \left( \begin{array}{c}  a+a \\ b+b  \end{array} \right) \otimes \left( \begin{array}{c}  c+e \\ d+f  \end{array} \right).$$
This gets even worse with the norm or things like this since the right norm is the product of the norms of the two vectors: 
$$ \left\| \left( \begin{array}{c}  a \\ b  \end{array} \right) \otimes \left( \begin{array}{c}  c \\ d  \end{array} \right) \right\| = \left\| \left( \begin{array}{c}  a \\ b  \end{array} \right) \right\|_1 \cdot \left\| \left( \begin{array}{c}  c \\ d  \end{array} \right) \right\|_2. $$
There is actually a way of rewriting a tensor product of two vectors  as one vector, but in a different way than you did. The tensor product of our spaces is four-dimensional, with a basis e.g. given as
$$ \left( \begin{array}{c}  1 \\ 0  \end{array} \right) \otimes \left( \begin{array}{c}  1 \\ 0 \end{array} \right), \left( \begin{array}{c}  1 \\ 0  \end{array} \right) \otimes \left( \begin{array}{c}  0 \\ 1 \end{array} \right), \left( \begin{array}{c}  0 \\ 1  \end{array} \right) \otimes \left( \begin{array}{c}  1 \\ 0 \end{array} \right), \left( \begin{array}{c}  0 \\ 1  \end{array} \right) \otimes \left( \begin{array}{c}  0 \\ 1 \end{array} \right) .$$
In this basis, the four-entry vector from your question correctly reads
$$ \left( \begin{array}{c}  a \alpha \\ a \beta \\ b \alpha \\ b \beta  \end{array} \right) $$
and now you can compute the norm the usual way:
$$ \sqrt{  a^2 \alpha^2 + a^2 \beta^2 + b^2 \alpha^2 + b^2 \beta^2 } = 1.$$
