I have a vector space $V$ and a subspace $W$ and it's complement $W'$ $(V = W+W',W\cap W' = \{0\})$
What does it mean that $\pi$ is the projection of $V$ unto $W$ which is parallel to $W'$?
Does it just mean that it's the projection of $V$ unto $W$? (the notion of parallel being implicit since $W'$ is complement to $W$)
Yes, any direct decomposition $V=W\oplus W'$ yields a projection $\pi$ to $W$ defined by $w+w'\mapsto w$.
The decomposition can be written up made of $\pi$, by $v=\pi(v)\,+\,(v-\pi(v))$.
Here the summand $\pi(v)$ is parallel to $W$ (actually $\pi(v)\in W$) and $v-\pi(v)$ is parallel to $W'$.
However, I would rather emphasize here that this projection is parallel to $W'$, meaning that $v-\pi(v)\in W'$ for all $v\in V$.
In the specific case (for Euclidean spaces) when $W'=W^\perp$, we talk about orthogonal projection, which acts parallelly to $W^\perp$.