Let $Y(j)$ be i.i.d. with finite mean and variance, and set $\mu=\mathbb{E}(Y)$ and $\tau=\sqrt{\mathbb{E}(Y^2)}$. If $(N(t))$ is an independent Poisson process with rate $\lambda$, then the compound Poisson process is defined as $X(t)=\sum_{j=0}^{N(t)} Y(j).$
The characteristic function of $X(t)$ is calculated as follows: for real $s$ we have \begin{eqnarray*} \psi(s)&=&\mathbb{E}\left(e^{is X(t)}\right)\cr &=&\sum_{j=0}^\infty \mathbb{E}\left(e^{is X(t)} \ | \ N(t)=j\right) \mathbb{P}(N(t)=j)\cr &=&\sum_{j=0}^\infty \mathbb{E}\left(e^{is (Y(1)+\cdots +Y(j))} \ | \ N(t)=j\right) \mathbb{P}(N(t)=j)\cr &=&\sum_{j=0}^\infty \mathbb{E}\left(e^{is (Y(1)+\cdots +Y(j))}\right) \mathbb{P}(N(t)=j)\cr &=&\sum_{j=0}^\infty \phi_Y(s)^j {(\lambda t)^j\over j!} e^{-\lambda t}\cr &=& \exp(\lambda t [\phi_Y(s)-1]) \end{eqnarray*} where $\phi_Y$ is the characteristic function of $Y$.
From this we easily calculate $\mu(t):=\mathbb{E}(X(t))=\lambda t \mu$ and $\sigma(t):=\sigma(X(t))= \sqrt{\lambda t} \tau$.
Take the expansion $\phi_Y(s)=1+is\mu -s^2\tau^2 /2+o(s^2)$ and substitute it into the characteristic function of the normalized random variable ${(X(t)-\mu(t)) /\sigma(t)}$ to obtain
\begin{eqnarray*} \psi^*(s) &=& \exp(-is(\mu(t)/\sigma(t))) \exp(\lambda t [\phi_Y(s/\sigma(t))-1]) \ &=& \exp(-s^2/2 +o(1)) \end{eqnarray*} where $o(1)$ goes to zero as $t\to\infty$. This gives the central limit theorem ${X(t)-\mu(t)\over\sigma(t)}\Rightarrow N(0,1).$
We may replace $\sigma(t)$, for example, with $\tau \sqrt{N(t)}$ to get ${X(t)-\mu(t)\over\tau \sqrt{N(t)}}= {X(t)-\mu(t)\over\sigma(t)} \sqrt{\lambda t \over N(t)} \Rightarrow N(0,1),$ by Slutsky's theorem, since $\sqrt{\lambda t \over N(t)}\to 1$ in probability by the law of large numbers.
Added: Let $\sigma=\sqrt{\mathbb{E}(Y^2)-\mathbb{E}(Y)^2}$ be the standard deviation of $Y$, and define the sequence of standardized random variables $T(n)={\sum_{j=1}^n Y(j) -n\mu\over\sigma\sqrt{n}},$ so that ${X(t)-\mu N(t)\over \sigma \sqrt{N(t)}}=T(N(t)).$
Let $f$ be a bounded, continuous function on $\mathbb{R}$. By the usual central limit theorem we have $\mathbb{E}(f(T(n)))\to \mathbb{E}(f(Z))$ where $Z$ is a standard normal random variable.
We have for any $N>1$, $\begin{eqnarray*} |\mathbb{E}(f(T(N(t)))) - \mathbb{E}(f(Z))| &=& \sum_{n=0}^\infty |\mathbb{E}(f(T(n)) - \mathbb{E}(f(Z))|\ \mathbb{P}(N(t)=n) \cr &\leq& 2\|f\|_\infty \mathbb{P}(N(t)\leq N) +\sup_{n>N} |\mathbb{E}(f(T(n)))- \mathbb{E}(f(Z)) |. \end{eqnarray*} $ First choosing $N$ large to make the right hand side small, then letting $t\to\infty$ so that $\mathbb{P}(N(t)\leq N)\to 0$, shows that $ \mathbb{E}(f(T(N(t)))) \to \mathbb{E}(f(Z)). $ This shows that $T(N(t))$ converges in distribution to a standard normal as $t\to\infty$.