Proving $L'=\{uv : u \in L \; \land v \in L^R \; \land |u|=|v|\}$ is a CFL with closure properties
Let $\tilde\Sigma = \{\tilde a,\tilde b\}$. It is well-known that the language $X = \{a^n b^n : n \geq 0\}$ is context-free. Let $h \colon \Sigma \cup \tilde\Sigma \to \Sigma$ be the homomorphism defined by $h(a) = h(b) = a$ and $h(\tilde a) = h(\tilde b) = b$. The following language is context-free:
$$ h^{-1}(X) = \{uv : u \in \Sigma^*, v \in \tilde\Sigma^*, |u| = |v| \}. $$
Let $t\colon \Sigma \to \tilde\Sigma$ be the homomorphism defined by $t(a) = \tilde a$, $t(b) = \tilde b$. The following language is regular:
$$
Lt(L)^R = \{ uv : u \in L, v \in t(L)^R \}.
$$
Since the context-free languages are closed under regular intersection, it follows that the following language is context-free:
$$
h^{-1}(X) \cap Lt(L)^R = \{ uv : u \in L, v \in t(L)^R, |u| = |v| \}.
$$
Finally, let $s\colon \Sigma \cup \tilde\Sigma \to \Sigma$ be the homomorphism defined by $s(a) = s(\tilde a) = a$, $s(b) = s(\tilde b) = b$. The following language is context-free:
$$
s(h^{-1}(X) \cap Lt(L)^R) = \{ uv : u \in L, v \in L^R, |u| = |v| \}.
$$
This is the language you're interested in.