If $P_{1}=\frac{\alpha_{1}\beta_{1}^{T}}{\alpha_{1}^{T}\beta_{1}}$ and $P_{2}=\frac{\alpha_{2}\beta_{2}^{T}}{\alpha_{2}^{T}\beta_{2}}$, then
$P_{1}P_{2}=\frac{\alpha_{1}\beta_{1}^{T}}{\alpha_{1}^{T}\beta_{1}}\frac{\alpha_{2}\beta_{2}^{T}}{\alpha_{2}^{T}\beta_{2}}$, where $P_{i},\;i=1,\;2$ are rank-1 projectione matrices.
However, if the situation like this
$\frac{\alpha_{2}^{T}\beta_{1}}{\alpha_{1}^{T}\beta_{1}}\frac{\alpha_{1}\beta_{2}^{T}}{\alpha_{2}^{T}\beta_{2}}$, then how to express the latter in terms of projection matrices?
Your two products are the same (assuming that $\alpha_1, \alpha_2, \beta_1, \beta_2$ are column vectors of the same size): that is, \begin{align} \frac{\alpha_{2}^{T}\beta_{1}}{\alpha_{1}^{T}\beta_{1}}\frac{\alpha_{1}\beta_{2}^{T}}{\alpha_{2}^{T}\beta_{2}} = \frac{\alpha_{1}\beta_{1}^{T}}{\alpha_{1}^{T}\beta_{1}}\frac{\alpha_{2}\beta_{2}^{T}}{\alpha_{2}^{T}\beta_{2}} . \label{darij1.eq.1} \tag{1} \end{align}
Proof. Forget about the denominators; they are just scalars and they are the same on both sides. Thus, it remains to prove that $\left(\alpha_2^T \beta_1\right) \left( \alpha_1 \beta_2^T\right) = \left(\alpha_1 \beta_1^T\right) \left(\alpha_2 \beta_2^T\right)$.
Now, let $\lambda$ be the scalar $\alpha_2^T \beta_1$; thus, to be fully precise, $\alpha_2^T \beta_1 = \lambda I_1$. Hence, $\lambda I_1 = \alpha_2^T \beta_1 = \beta_1^T \alpha_2$ (since $x^T y = y^T x$ for any two column vectors $x$ and $y$). Now, \begin{align} \underbrace{\left(\alpha_2^T \beta_1\right)}_{=\lambda} \left( \alpha_1 \beta_2^T\right) &= \lambda \left(\alpha_1 \beta_2^T\right) = \alpha_1 \underbrace{\lambda I_1}_{= \beta_1^T \alpha_2} \beta_2^T = \alpha_1 \left(\beta_1^T \alpha_2\right) \beta_2^T = \left(\alpha_1 \beta_1^T\right) \left(\alpha_2 \beta_2^T\right) . \end{align} And this is precisely what we needed to prove.