This one is a bit long to put as a reply, so I give it as another answer

I agree with Pinelis that the conditional version of CLT would be required if we want to deduce the following statement:

$$\mathbb{E}\left( f\left( \frac{S_n}{\sqrt{2n}} \right) \left\vert \frac{S_{2n}}{\sqrt{2n}} =\frac{[X\sqrt{2n}]}{\sqrt{2n}} \right. \right) \xrightarrow{\text{in laws}} \mathbb{E}\left( f\left( \frac{G_1+G_2}{2} \right) \vert G_2=X \right)\text{ (*) } $$

where $X$ is a bounded random variable, $(G_1,G_2)$ is a centered reduced gaussian vector.
However, as I understand our problem is to estimate the following quantity $$ \text{limit of } \mathbb{P}\left( \left|\frac{S_n}{\sqrt{2n}}\right| \le a \left| \left| \frac{S_n}{\sqrt{2n}} \right| \le b \right. \right) =L(a,b)$$

for 2 positive constants $a,b$, and to answer if
$L(a,b) \longrightarrow 0 $ when $ a \longrightarrow \infty \text{ (**) }$

And indeed, the CLT and the functional version of Portementeau's theorem gives us the reply, as I mentioned.
More precisely, we are in fact considering an (a Lesbeque) almost everywhere continous, and by Portementeau, we have that :

$$L(a,b) = \mathbb{P}\left( \frac{|G_1+G_2|}{2} \le a \left\vert |G_2| \le b \right.\right)$$

This one is much better than what is required to answer (*).

Remark: In fact, if my calculation is correct, (**) is indeed true in the case $U$ follows the uniform law on any interval. Its proof is rather technique, but elementary.

This one is a bit long to put as a reply and it is also a quantitative reply for the given problem, so I give it as another answer

I agree with Pinelis that the conditional version of CLT would be required if we want to deduce the following statement:

$$\mathbb{E}\left( f\left( \frac{S_n}{\sqrt{2n}} \right) \left\vert \frac{S_{2n}}{\sqrt{2n}} =\frac{[X\sqrt{2n}]}{\sqrt{2n}} \right. \right) \xrightarrow{\text{in laws}} \mathbb{E}\left( f\left( \frac{G_1+G_2}{2} \right) \vert G_2=X \right)\text{ (*) } $$

where $X$ is a bounded random variable, $(G_1,G_2)$ is a centered reduced gaussian vector.
However, as I understand our problem is to estimate the following quantity $$ \text{limit of } \mathbb{P}\left( \left|\frac{S_n}{\sqrt{2n}}\right| \le a \left| \left| \frac{S_n}{\sqrt{2n}} \right| \le b \right. \right) =L(a,b)$$

for 2 positive constants $a,b$, and to answer if
$L(a,b) \longrightarrow 0 $ when $ a \longrightarrow \infty \text{ (**) }$

And indeed, the CLT and the functional version of Portementeau's theorem gives us the reply, as I mentioned.
More precisely, we are in fact considering an (a Lesbeque) almost everywhere continous, and by Portementeau, we have that :

$$L(a,b) = \mathbb{P}\left( \frac{|G_1+G_2|}{2} \le a \left\vert |G_2| \le b \right.\right)$$

This one is much better than what is required to answer (**).

Remark: In fact, if my calculation is correct, (*) is indeed true in the case $U$ follows the uniform law on any interval. Its proof is rather technique, but elementary.