Projection onto a subspace

Linearity. For $x,y\in\mathbb{R}^n$ and $\lambda\in\mathbb{R}$:
$$p(x+\lambda y)=\sum_i\big(\langle x,u_i\rangle+\lambda\langle y,u_i\rangle\big)u_i=p(x)+\lambda p(y).$$

Image. $p(x)$ is a combination of the $u_i$, so $\operatorname{Im} p\subseteq F$. Conversely, if $x\in F$, its expansion in the orthonormal basis is $x=\sum_i\langle x,u_i\rangle u_i=p(x)$, so $x\in\operatorname{Im} p$. Hence $\operatorname{Im} p=F$, and $p$ is the identity on $F$.

Kernel. The $u_i$ are linearly independent (orthonormal), so $p(x)=0\iff\langle x,u_i\rangle=0$ for all $i$. Since the $u_i$ span $F$, this holds iff $x\perp F$. Hence $\ker p=F^\perp$.

Idempotence. $p(x)\in F$ and $p$ is the identity on $F$, so $p(p(x))=p(x)$.

Conclusion. $p$ is the orthogonal projection onto $F$: $\operatorname{Im} p=F$, $\ker p=F^\perp$, $p^2=p$. Moreover $x=p(x)+(x-p(x))$ with $p(x)\in F$ and $x-p(x)\in\ker p=F^\perp$ (because $p(x-p(x))=p(x)-p^2(x)=0$), so $\mathbb{R}^n=F\oplus F^\perp$.