Let $V$ be an inner product space and $v, u_1, u_2\in V$ where $(u_1, u_2)$ are linearly independent.
Define $U=Span(u_1, u_2)$ and $$w_1 = \operatorname{proj}_{u_1}(v) \\ w_2 = \operatorname{proj}_{u_2}(v) \\ w = \operatorname{proj}_{U}(v)$$
Is it necessarily true that $w=w_1+w_2$?
I think that it's not true, but I can't find any counter-example! because it's seem that there are no relation to the sum of them and the fact given on $v$.
Let's take $V = \mathbb{R}^3$ and $u_1 = (1,0,0), u_2 = (1,1,0)$. Then $U = \operatorname{Span} \{ u_1, u_2 \}$ is the $xy$ plane and the orthogonal projection of a vector $v = (x,y,z)$ on $U$ is given by $w = (x,y,0)$. The orthogonal projection of $v$ on $u_1$ is
$$ w_1 = \frac{\left< w, u_1 \right>}{\left< u_1, u_1 \right>} u_1 = (x,0,0) $$
while the orthogonal projection of $v$ on $u_2$ is
$$ w_2 = \frac{\left< w, u_2 \right>}{\left< u_2, u_2 \right>} u_2 = \frac{x + y}{2} (1, 1, 0) $$
and
$$ w_1 + w_2 = \left( \frac{3x + y}{2}, \frac{x + y}{2}, 0 \right) \neq w. $$
Note that the statement is true if $u_1,u_2$ form an orthogonal basis of $U$ so if instead of our $u_1,u_2$ we would take $u_1 = (1,0,0)$ and $u_2 = (0,1,0)$ then everything would work.