Each coin falls head in the second round if and only it fell head in the first round (hence was tossed again) and fell head in the second round as well. Thus each coin falls head in the second round with probability $p^2$, and the number $R_2$ of these has binomial $(n,p^2)$ distribution.
Since $R_1$ is binomial $(n,p)$, $\mathbb E(R_1)=np$, $\mathrm{var}(R_1)=np(1-p)$ and $\mathbb E(R_1^2)=\mathrm{var}(R_1)+\mathbb E(R_1)^2=np(1-p)+n^2p^2$. Since $R_2$ is binomial $(n,p^2)$, $\mathbb E(R_2)=np^2$ and $\mathrm{var}(R_2)=np^2(1-p^2)$. For every $k$, conditionally on $R_1=k$, $R_2$ is binomial $(k,p)$, hence $\mathbb E(R_2\mid R_1)=R_1p$, and $ \mathbb E(R_1R_2)=\mathbb E(R_1\mathbb E(R_2\mid R_1))=p\mathbb E(R_1^2). $ Thus, $ \text{Cov}(R_1,R_2)=\mathbb E(R_1R_2)-\mathbb E(R_1)\mathbb E(R_2)=np^2(1-p), $ and $ \text{Corr}(R_1,R_2)=\frac{\text{Cov}(R_1,R_2)}{\sqrt{\text{var}(R_1)\text{var}(R_2)}}=\sqrt{\frac{p}{1+p}}. $