Probably the most important difference between the swap test and the projective measurement technique is that performing the swap test does not require knowledge of its inputs.
When applying the swap test, once you are provided the states $|\psi\rangle, |\phi\rangle$, you need no additional information about the systems to estimate $|\langle \psi | \phi \rangle|^2$.
With the projective measurement technique, you need to implement a measurement $\{|\phi\rangle\langle \phi|, I - |\phi\rangle\langle \phi|\}$ in order to estimate $|\langle \psi | \phi \rangle|^2$, which involves having some knowledge about $|\psi\rangle$ or $|\phi\rangle$ or how the states are prepared. For example, this measurement is easy to perform is if you are provided $|\psi\rangle$ and also given the ability to implement a unitary $V^\dagger$ where $V$ satisfies $|\phi\rangle = V |0\rangle$. In this case, you define your measurements as
\begin{align}
|\phi\rangle\langle \phi| &= V |0\rangle \langle 0| V^\dagger \tag{1} \\
I - |\phi\rangle\langle \phi| &= V \left(I - |0\rangle \langle 0|\right) V^\dagger \tag{2}
\\&= V\left(\sum_{x\in\{0,1\}^n\backslash0} |x\rangle \langle x|\right)V^\dagger\tag{3}
\end{align}
This form of the measurements highlights that after applying $V^\dagger$ to $|\psi\rangle$, the projectors represent computational basis measurements where the probability of observing "0" is $p(0) = |\langle \psi | \phi \rangle |^2$. But again, this relied on the ability to implement $V^\dagger$, which is not guaranteed.
