$\mathcal{G} = \{ v_2 v_4 \ldots v_{k} ~:~ v_1 v_2 v_3 v_4 \ldots v_{k-1} v_{k} \in \mathcal{L}, ~ \text{k even} \} $ is context free language Let $\mathcal{L}$ be context free language over alphabet $\Sigma$. Define $\mathcal{G}$ as
$$\mathcal{G} = \{ v_2 v_4 \ldots v_{k} ~:~ v_1 v_2 v_3 v_4 \ldots v_{k-1} v_{k} \in \mathcal{L}, ~ \text{k even} \} $$
I have seen similar question (asked 5 years ago) but I am not sure how it can works.
Proposition
$\mathcal{L}$ is CFL so it has own push-down automata. So let copy states of $\mathcal{L}$ and if it has a state called $S$ and it gets to state $T$ upon letter $x$ then $\mathcal{G}$ will have states $S_1, S_2, T_1, T_2$ and letter $x$ turns $S_1$ to $T_2$ and $S_2$ to $T_1$.
My question is why it is correct apporach? $\mathcal{G}$ automata won't read any of $v_1, v_3, v_5,... v_{k-1}$ so how it can ensure that this word belongs to $\mathcal{L}$?
 A: Remember that PDAs can be nondeterministic, meaning they can "guess" which production rule to apply and, if any sequence of guesses ends up producing the word, the machine succeeds.
Here's how you can design a nondeterministic PDA to recognize $\mathcal{G}$. $\mathcal{G}$ is a modified version of the nondeterministic PDA that recognizes $\mathcal{L}$. First, it has two copies of its internal finite control states—call them red states and blue states.
If it is in a red state, then it behaves like the original $\mathcal{L}$ machine—it reads a symbol off the input and off the stack, then transitions normally. (But red states always transition into blue states.)
If it is in a blue state, then instead of reading the input, it nondeterministically picks a symbol $\alpha$, then makes the transition that $\mathcal{L}$ would have done if it had just read $\alpha$ off of the input tape.  (But blue states always transition into red states.)
That's the missing piece. The machine alternates between red states that actually read an input symbol and transition the way the $\mathcal{L}$ machine does, and blue states that nondeterministically guess a symbol and simulate what the $\mathcal{L}$ machine does.
In this way, the overall machine takes in a word, nondeterministically guesses the missing characters that occur every other letter, and accepts if the overall guessed word is in $\mathcal{L}$. This is how it recognizes the "half" words in $\mathcal{G}$.
