If $R$ is a unital ring and $e$ is an idempotent element, i.e. $e^2=e$, we have that also $(1-e)$ is idempotent. Morover if $e$ is in the centre of $R$ is easy to show that $R \cong R/(e) \ \times R/(1-e)$. The ideal generated by $(e)$ and $(e-1)$ contains $e + (1-e)=1$, thus it is $R$. Whilts if $a$ is in their intersection we have $a=r_1e=r_2(1-e)$, so $0=r_2(1-e)e=r_1e^2=r_1e=a$, so the intersection is trivial (here we have used the fact that $e$ and $e-1$ are in the centre). Therefore $(e)$ and $(1-e)$ are comaximal.
Let $V$ be a vector space on $F$ and $End(V)$ his ring of endomorphisms with, as usual, the pointwise addiction as the sum and the composition as multiplication. Its idempotent elements, called projections, don't need to be in the centre. For example
$A=\begin{pmatrix} 1 & 0 \\ 1 & 0\end{pmatrix}$
clearly $A^2=A$ but
$\begin{pmatrix} 1 & 0 \\ 1 & 0\end{pmatrix}\begin{pmatrix} a & b \\ c & d\end{pmatrix} = \begin{pmatrix} a & b \\ a & b\end{pmatrix}$, meanwhile
$\begin{pmatrix} a & b \\ c & d\end{pmatrix} \begin{pmatrix} 1 & 0 \\ 1 & 0\end{pmatrix}= \begin{pmatrix} a+b & 0 \\ c+d & 0\end{pmatrix}$
But, from the reading I'm doing, the result should hold also in this setting. So I'm looking for a general proof that if $\theta$ is idempotent in $End(V)$ then $(\theta)$ and $(I-\theta)$ are comaximal (or a disprove). Any help would be appreciated, thank you.
The statement is wrong. The ring $End(V)$ is simple so the ideals $(\theta)$ and $( 1-\theta)$ are equal to $End(V)$ and are comaximal iff $\theta=0$ or $\theta=1$.