Assume we have a subspace $V$ of $\mathbb{R}^n$ and a given vector $\vec x$ not in $V$.
If $B \subseteq V$ is an orthonormal basis of $V$ then we have:
$$proj_V(\vec{x}) = \sum^n_{i=1}proj_{u_i}(\vec{x})$$
Where $\{u_i\} = B$.
Why must we have an orthonormal Basis? Why isn't it enough to have an arbitrary basis, and why isn't it enough to have an orthogonal basis?
This isn't strictly necessary, any basis can be protected onto.
The usefulness of an orthonormal basis comes from the fact that each basis vector is orthogonal to all others and that they are all the same "length".
Consider the projection onto each vector separately, which is "parallel" in some sense to the remaining vectors, so it has no "length" in those vectors. This means you can take the projection onto each vector separately and then add them to get the overall projection onto the basis, knowing that none of the individual projections will "interfere" with any component of the others.
Given that each basis vector has unit length, there is no scaling needed to complete this addition of individual projections either, so further calculation is eliminated.