Theorem 10: If $\{u_1,...,u_p\}$ is an orthonormal basis for a subspace $W$ of $\mathbb{R}^n$, then $proj_Wy= (y \cdot u_1)u_1 + ... + (y \cdot u_p)u_p$. If $U =[u_1 \ \ \ u_2 \ \ \ u_p]$ ,then $proj_W y = UU^Ty$ for all $y$ in $\mathbb{R}^n$
$proj_Wy= (y \cdot u_1)u_1 + ... + (y \cdot u_p)u_p$ shows that $proj_Wy$ is a linear combination of the columns of $U$ using the weights $y \cdot u_1 + ... + y \cdot u_p$ The weights can be written as $u^T_1y,...,u^T_py$ showing that they are the entries in $U^Ty$.
I don't see how $y \cdot u_1 + ... + y \cdot u_p$ is obtained. If we distribute the $u_i$ in the equation, wouldn't it be $proj_Wy= yu + ... + yp$?
It is the definition of the projection, that is how it is defined. But here is a bit of intuition behind the definition:
Suppose you have $2$ vectors in 3D space, then their inner product is defined like this:
Notice that the length of the projection of $\vec{a}$ on $\vec{b}$ is $$|\vec{a}_{proj}| = |\vec{a}|\cdot \cos \theta = \frac{\vec{a}\cdot \vec{b}}{|\vec{b}|} $$ Maybe a picture will help:
$\vec{a}_{proj}$ points at same direction of $\vec{b}$ so if $|\vec{b}|=1$, meaning $b$ is a unit vector, then $$\vec{a}_{proj} = |\vec{a}_{proj}| \vec{b} = \frac{\vec{a}\cdot \vec{b}}{|\vec{b}|} \vec{b} = (\vec{a}\cdot \vec{b}) \vec{b} $$
Now, if you want to project a vector $\vec{a}$ on a plane spanned by the orthonormal basis $\{\vec{b_1},\vec{b_2}\}$. just project the vector on each of the vectors in the basis and add the projections.