I can see how the equivalence can formulated with
(P -> R) V (Q -> R)
= (¬P V R) V (¬Q V R)
= (¬P ∧ ¬Q) V R
= ¬(P ∧ Q) V R
= (P ∧ Q) -> R
(Sorry, I would've written this in LaTeX if I were more competent.)
although I still it counter-intuitive and, at a glance, first thought it was...