Can the regular image of a context-free language be undecidable?
If $L_1$ is context-free, then so is $L_2$. You can show this easily using closure properties of context-free languages. Let $\Sigma' = \{ \sigma' : \sigma \in \Sigma \}$; we assume that $\Sigma$ and $\Sigma'$ are disjoint. Define a homomorphism $h\colon \Sigma \cup \Sigma' \cup \{|\} \to \Sigma^* \cup \{|\}$ by $h(\sigma) = h(\sigma') = \sigma$ and $h(|) = |$, and a homomorphism $k\colon \Sigma \cup \Sigma' \cup \{|\} \to \Sigma^*$ by $k(\sigma) = \sigma$ and $k(\sigma') = k(|) = \epsilon$. Then
$$
L_2 = k(h^{-1}(L_1) \cap \Sigma^{\prime\ast} | \Sigma^*).
$$
The family of context-free languages is closed under homomorphisms, inverse homomorphisms, and intersection with regular languages, hence $L_2$ is context-free as well.