I'm trying to understand why the statement above is true for a finite dimensional inner product space V.
After some research I found that as long as $P$ is a linear operator on V and is idempotent, then it is true that V is the direct sum of the range and nullspace of $P$.
Then, I also know that for any $v\in V$, $v=u+w$ for some $u\in\text{null}(P)$ and $w\in\text{Im}(P)$, and using the fact that $P=P^2$, we have $$Pv=P(u+w)=Pw=w$$
Furthermore, I know that $P$ is an orthogonal projection of $V$ onto some subspace $W$ of $V$ if
$Pw=w\text{ for all }w\in W\text{ and}$
$Pw'=0\text{ for all }w'\in W^\perp$
Then, all that is left to show is that $\text{null}(P)=(\text{Im}(P))^\perp$.
Finally, I also know that if $V$ is finite dimensional, $V=\text{Im}(P)\oplus(\text{Im}(P))^\perp$.
So then my question is: Does it immediately follow that $\text{null}(P)=(\text{Im}(P))^\perp$? And if it does, why does the nullspace of $P$ have to be a subset of the orthogonal complement of its image?
The answer for your first question is no. Take $P=\begin{pmatrix}1 & 1 \\ 0 & 0\end{pmatrix}$ then it is idempotent but is not an orthogonal projection matrix. Here, $\text{null}(P) = \text{span}\{ (1,-1) \}$ whereas $(\text{Im}(P))^\perp = \text{span}\{ (0,1) \}$.
However, if we assume $\text{null}(P)\subseteq(\text{Im}(P))^\perp$ then by the decomposition $$V=\text{Im}(P)\oplus(\text{Im}(P))^\perp = \text{Im}(P)\oplus \text{null}(P)$$ we know that $\dim(\text{null}(P))=\dim((\text{Im}(P))^\perp)$ so it follows that $$\text{null}(P)=(\text{Im}(P))^\perp$$ Hence, $P$ is an orthogonal projection matrix.