As Demetri says, the answer is right there - maybe an example helps?

Consider $f(1|\theta ,y)=P(y_{f}=1|\theta)$, assuming random sampling. Note $P(y_{f}=1|\theta)=E(y|\theta)=\theta$. Hence, \begin{eqnarray} f(y_{f}=1|y)&=&\int f(y_{f}=1|\theta)\pi(\theta|y)d\theta\notag\\ &=&\frac{\Gamma\left(\alpha _{0}+\beta_{0}+n\right)}{\Gamma\left(\alpha_{0}+k\right)\Gamma\left(\beta_{0}+n-k\right)}\int \theta\theta ^{\alpha _{0}+k-1}\left( 1-\theta \right) ^{\beta _{0}+n-k-1}d\theta\notag\\ &=&\frac{\Gamma\left(\alpha_{0}+\beta_{0}+n\right)}{\Gamma\left(\alpha_{0}+k\right)\Gamma\left(\beta_{0}+n-k\right)}\frac{\Gamma\left(\alpha_{0}+k+1\right)\Gamma\left(\beta_{0}+n-k\right)}{\Gamma\left(\alpha_{0}+\beta_{0}+n+1\right)}\notag\\ &=&\frac{\alpha_{0}+k}{\alpha _{0}+\beta_{0}+n}, \end{eqnarray} using $\Gamma(\alpha)=(\alpha-1)\Gamma(\alpha-1)$.

As Demetri says, the answer is right there - maybe an example helps?

Consider $f(y^{new}=1|\theta ,y)=P(y_{f}=1|\theta)$, i.e., the probability that the next attempt will be a success, assuming random sampling. Note $P(y_{f}=1|\theta)=E(y|\theta)=\theta$. 

Hence, with $m$ and $k$ the number of attempts and successes in the first sample, \begin{eqnarray} p(y^{new}=1|y)&=&\int f(y^{new}=1|\theta)\pi(\theta|y)d\theta\notag\\ &=&\frac{\Gamma\left(\alpha _{0}+\beta_{0}+n\right)}{\Gamma\left(\alpha_{0}+k\right)\Gamma\left(\beta_{0}+n-k\right)}\int \theta\theta ^{\alpha _{0}+k-1}\left( 1-\theta \right) ^{\beta _{0}+n-k-1}d\theta\notag\\ &=&\frac{\Gamma\left(\alpha_{0}+\beta_{0}+n\right)}{\Gamma\left(\alpha_{0}+k\right)\Gamma\left(\beta_{0}+n-k\right)}\frac{\Gamma\left(\alpha_{0}+k+1\right)\Gamma\left(\beta_{0}+n-k\right)}{\Gamma\left(\alpha_{0}+\beta_{0}+n+1\right)}\notag\\ &=&\frac{\alpha_{0}+k}{\alpha _{0}+\beta_{0}+n}, \end{eqnarray} using $\Gamma(\alpha)=(\alpha-1)\Gamma(\alpha-1)$.