Prove that a sum of projections is a projection iff they are orthogonal, if the characteristic of the space is not $2$

Easy direction: If $E_1E_2=E_2E_1=0$, then $(E_1+E_2)^2=E_1^2+E_2^2 = E_1+E_2$.

Conversely, suppose $(E_1+E_2)^2= E_1+E_2$. Then $$ E_1E_2+E_2E_1 = 0 \tag{1} $$ By multiplying both sides of (1) by $E_1$ on the left, or by $E_1$ on the right, obtain two equalities: $$E_1 E_2+E_1E_2E_1=0,\qquad E_1E_2E_1+E_2E_1=0$$ By subtracting these, $$E_1E_2-E_2E_1=0. \tag{2}$$ Adding or subtracting (1) and (2), we get $$2E_1E_2=0,\qquad 2E_2E_1=0$$ Since the characteristic is not $2$, the factor $2$ can be cancelled.

(This answer is based on this post by Alex).

Think about it this way: $E_1+E_2$ is a projection if it satisfies: $(E_1+E_2)^2=(E_1+E_2)$

(Use $E_iE_j$ to mean the composition)

1)Assume $E_1E_2=0$

We want to show that $(E_1+E_2)(E_1+E_2)=(E_1+E_2)$ This means that $E_2E_1+E_2E_2+.....=(E_1+E_2)$ Can you see the next step?

For the converse, assume $(E_1+E_2)$ is a projection, then it must satisfy $(E_1+E_2)^2=....$