Let $u_n$ be a measure having uniform distribution on {$-1,1$}$^n$ (elements of {$-1,1$}$^n$ are $n$ dimensional vectors with values in {$-1,1$}) . Let $v$ be a measure having standard gaussian distribution in one dimension. Let $g_n(x_1,...,x_n)=f(\frac{x_1+...+x_n}{\sqrt{n}})$. In the book I am reading author says that
$$\lim_{n \to \infty}\int g_n^2log(g_n^2)$d_{u_n}-\int(g_n)^2d_{u_n}log(\int (g_n)^2d_{u_n}=\int f^2logf^2d_{v}-\int f^2d_{v}log(\int f^2d_{v})$$
due to the central limit theorem.
Central limit theorem in its one version states that if $X_1,...,X_n$ are independent random variables with same distribution, 0 mean and 1 variances then $\frac{X_1+...+X_n}{\sqrt{n}}$ converges in distribution to $X$ which has standard normal distribution.
$Y_n$ converge in distribution to $Y$ iff $\lim_{n \to \infty} \int f d_{u_n} = \int f d_u$ (here the measures are not related to measures from question)
I can't connect these definition to statement from the book. In book we have sequence of functions $g_n$ instead of just a function $f$ as in definition of convergence in distribution. Moreover we don't have a sequence of random variables yet.
Inutitively I can feel why its true but I don't have mathematical argument. Can you help me?
The argument is very simple if you assume $f$ is continuous and bounded. Otherwise you'd need more than convergence in distribution and the central limit theorem to get this result.
Let $h: \mathbb{R} \to \mathbb{R}$ be any continuous function. Let $U_1,U_2,\dots,$ be an infinite sequence of i.i.d. Rademacher (uniform over $\{0,1\}$) random variables and let $X$ be a standard Gaussian random variable.
By the central limit theorem, $\frac{U_1 + \dots + U_n}{\sqrt{n}}$ converges weakly to $X$. By the definition of weak convergence, $$\mathbb{E}\left[s\left(\frac{U_1 + \dots + U_n}{\sqrt{n}}\right)\right] \to \mathbb{E}[s(X)]$$ for any bounded, a.e. continuous function $s$ ($s$ only needs to be a.s. continuous with respect to $X$, but since $X$ is a Gaussian this is the same as being a.e. continuous). Because $h$ is continuous and $f$ is bounded and continuous, $h\circ f$ is a bounded, continuous function. Combining everything together,
$$\int h(g_n(x_1,\dots,x_n))\,d_{u_n} = \int h\left(f\left(\frac{x_1 + \dots+x_n}{\sqrt{n}}\right)\right)\,d_{u_n} =\mathbb{E}\left[h\circ f\left(\frac{U_1 + \dots + U_n}{\sqrt{n}}\right)\right] \to \mathbb{E}[h\circ f(X)] = \int h(f)\,dv.$$
You can apply this to your equations by setting $h(x) = x^2$ and $h(x) = x^2\log(x^2)$. Note that the second function is technically ill-defined at $0$, but $\mathbb{P}(X = 0) = 0$ so we can ignore this discontinuity.
If $f$ can be discontinuous or if $f$ is unbounded then you would need to do a little more work showing that the limit still holds.