I'll be going in a circle while trying to answer the question:
First of all, given an orthonormal basis $\{e_n\}_{n\in \mathbb{N}}$ of a (separable infinite-dimensional) Hilbert space, Parseval's identity—the equality case of Bessel's inequality—implies that $x = \sum_{n \in \mathbb{N}} \langle x, e_n\rangle e_n$, where the sum converges in norm.
The equality $\sum_{n\in\mathbb{N}} |\langle x, e_n\rangle|^{2} = \|x\|^2$ can be seen as an infinite-dimensional Pythagorean theorem.
Now if $\{f_n\}_{n \in \mathbb{N}}$ is an orthonormal system, the vector $Px = \sum_{n\in\mathbb{N}} \langle x,f_n\rangle f_n$ lies in the closed linear span of $\{f_n\}_{n}$ and its squared length is $\|Px\|^2 = \sum_{n\in\mathbb{N}} |\langle x,f_n\rangle|^2$ by Parseval. Now Bessel's inequality tells us that $\|Px\| \leq \|x\|$ and thus the map $x \mapsto Px$ is continuous of norm $1$ since $P$ is obviously linear. It satisfies $P^2 = P$ and it is easy to check that $P^{\ast} = P$, so we see that $P$ is an orthogonal projection and by definition its range is the closed span of $\{f_n\}$. Obviously, $x = Px + (1-P)x$ and $P^{\ast} = P$ gives us that $\langle Px, (1-P)x \rangle = \langle x, (P-P^2)x \rangle = 0,$ so $Px \perp (1-P)x$. Now in the (at most) two-dimensional subspace spanned by $Px$ and $(1-P)x$ we have a right-angled triangle with corners $0,Px,x$. By expanding the scalar product, we see
$\|x\|^2 = \langle Px + (1-P)x, Px + (1-P)x \rangle = \|Px\|^2 + \|(1-P)x\|^2$ which is of course nothing but Pythagoras.
Now this equality together with $\|Px\|^2 = \sum_{n\in\mathbb{N}} |\langle x,f_n\rangle|^2$ gives us back Bessel's inequality
$\sum_{n\in\mathbb{N}} |\langle x,f_n\rangle|^2 \leq \|x\|^2$
and we see that the defect is measured by $(1-P)x$, the part of $x$ orthogonal to the span of the $\{f_n\}$.
Finally, the angle $\phi$ between $x$ and $Px$ is defined by $\cos{\phi} = \frac{\langle x, Px \rangle}{\|x\|\,\|Px\|},$ where the fraction is in $[-1,1]$ by Cauchy-Schwarz, and that gives us the interpretation of $Px$ as the "cosine projection" of the hypotenuse $x$ of our triangle with corners $0,Px,x$.
I'm not aware of a third thing that you missed (except if you're willing to count Parseval). I hope this answers your question, I for one found these considerations illuminating when I made them for the first time.