What trace properties are used in the identity ${\rm tr}_A{\rm tr}_B(\rho\Pi)={\rm tr}_A(\rho_A{\rm tr}_B(\rho_B\Pi))$?

To turn the probability of the projection over the Hilbert space $$\mathcal H_A \otimes \mathcal H_B$$ into the POVM probabilty over $$\mathcal H_A$$ we we use this equality: $$tr_Atr_B(ρΠ_i)=tr_A(ρ_Atr_B(ρ_BΠ_i))$$

My syllabus states that $$\rho=\rho_A\otimes \rho_B$$ but says nothing about the projector $$\Pi_i$$, as far as I know this is just an operator that can't necessarily be split into a tensor product and works on the full Hilbert space.

How does one obtain this equality?

What does $$tr_B(\rho_B \Pi_i)$$ even mean? As $$\rho_B$$ and $$\Pi_i$$ are matrices of different sizes.

What I have tried:
If I do assume that $$\Pi_i=\Pi_A\otimes \Pi_B$$ I get: \begin{align} tr_Atr_B(ρΠ_i)&= tr_Atr_B((\rho_A \otimes \rho_B)\cdot(\Pi_A\otimes \Pi_B))\\ &=tr_Atr_B((\rho_A\cdot \Pi_A)\otimes(\rho_B\cdot \Pi_B)\\ &=tr_A((\rho_A\cdot \Pi_A) tr_B(\rho_B\cdot \Pi_B) \end{align} Which is almost correct but now there is this $$\Pi_A$$ term. I'm not sure if I can move this $$\Pi_A$$ back into the $$tr_B$$ somehow. This is also all assuming I can even decompose the projector as a tensor product.

You can use the identity $$\mathrm{tr}_{B}\left( (Y_A \otimes \mathbb{I}_{B}) X_{AB} \right) = Y_A \mathrm{tr}_B(X_{AB})$$ where $$X_{AB}$$ can be any operator acting on the joint system $$AB$$. You can prove this quite quickly by writing the operators in the standard basis $$Y_A = \sum_{i,j} y_{ij} |i \rangle \langle j |$$ and $$X_{AB} = \sum_{a,b,c,d} x_{abcd} |a \rangle \langle b| \otimes |c \rangle \langle d|$$ and showing the two expressions are equal.
Using that expression you have \begin{aligned} \mathrm{tr}_A (\mathrm{tr}_B ( \rho \Pi)) &= \mathrm{tr}_A(\mathrm{tr}_B((\rho_A \otimes \rho_B) \Pi_{AB})) \\ &= \mathrm{tr}_A(\mathrm{tr}_B((\rho_A \otimes \mathbb{I}_B)(\mathbb{I}_A \otimes \rho_B) \Pi_{AB})) \\ &= \mathrm{tr}_A(\rho_A\mathrm{tr}_B((\mathbb{I}_A\otimes \rho_B)\Pi_{AB})) \end{aligned} which is what you wanted. Note that $$\rho_B \Pi_{AB}$$ is usually shorthand for $$(\mathbb{I}_A \otimes \rho_B) \Pi_{AB}$$.