Distinguish between two possible configurations for angular momentum in carbon atom
You are separating the spatial and spin part of your two wavefunctions. I don't think that is a good idea here, because it does only work for one of the two configurations.

The wavefunction of a configuration is given by a Slater determinant build up from spin orbitals. In the following I will neglected the normalization constant of $\frac{1}{\sqrt2}$ for simplicity. The wavefunction for the first configuration ($\uparrow - \downarrow$) is

\begin{align}
   \Psi_1 &amp;= \begin{vmatrix}
     p_-(1)\alpha(1) &amp; p_+(1)\beta(1)\\
     p_-(2)\alpha(2) &amp; p_+(2)\beta(2)\\
     \end{vmatrix}\\
    &amp;= p_-(1)\alpha(1)p_+(2)\beta(2) - p_+(1)\beta(1)p_-(2)\alpha(2)
\end{align}

Note that this differs from your suggested wavefunction by the spin part of the orbitals, and there is no way to separate them out.

And for the other configuration ($-\uparrow\downarrow -$) we have
\begin{align}
   \Psi_2 &amp;= \begin{vmatrix}
     p_0(1)\alpha(1) &amp; p_0(1)\beta(1)\\
     p_0(2)\alpha(2) &amp; p_0(2)\beta(2)\\
     \end{vmatrix}\\
    &amp;= p_0(1)\alpha(1)p_0(2)\beta(2) - p_0(1)\beta(1)p_0(2)\alpha(2)
\end{align}

Both, $\Psi_1$ and $\Psi_2$ look quite similar now. However, since the spatial part of both orbitals in $\Psi_2$ is the same here, we can factor them out

\begin{equation}
  \Psi_2 = p_0(1)p_0(2)\underbrace{[\alpha(1)\beta(2)-\beta(1)\alpha(2)]}_{\chi}
\end{equation}

And we get what you have suggested for this configuration. But this does only work for $\Psi_1$!

Now, why is $\Psi_1$ assigned to $L=2$, while $\Psi_2$ is $L=0$?

I assume your confusion is because if you add up the orbital quantum numbers $m_l$ you get $0$ in both cases ($1-1=0$ and $0+0=0$). However, those sums give you $M_L$, not $L$. The many-electron quantum number $L$ has $2L+1$ components with quantum numbers $M_L=-L,-L+1,\dots 0, \dots L-1,L$. So for $L=2$ we have $M_L=-2,-1,0,1,2$, much like there are three orbitals $p_-$, $p0$ and $p+$ for $l=1$

So $\Psi_1$ would be the $M_L=0$ component of $L=2$, while $\Psi_2$ is the $M_L=0$ component of $L=0$.

But there is one issue with such assignments: You cannot really do it, as the configurations get mixed together. The point is, that the atom as a whole needs to be spherical symmetric (unless there is some external electric or magnetic field). But the individual $p$ orbitals are not. For example $p_0$ is (in the usual convention) aligned along the $z$-axis and zero in the $xy$ plane.

Mathematically speaking the configurations form a basis (Hilbert space) in which the actual electronic states are expanded. This is known as the configuration interaction method.

So for example the electronic state with $L=0$ would be
\begin{equation}
\Psi_2 = \frac{1}{\sqrt3}(|\uparrow\downarrow - -\rangle + |-\uparrow\downarrow - \rangle + |--\uparrow\downarrow \rangle)
\end{equation}

where each $|\dots\rangle$ denotes a Slater determinant.