You want to show that two things happen:
- Every element of $V$ can be written as a sum $u+w$, where $u\in\mathrm{ker}(f-\mathrm{id}_V)$ and $v\in\mathrm{ker}(f+\ker{id}_v)$; and
- The intersection of $\mathrm{ker}(f-\mathrm{id}_V)$ and $\mathrm{ker}(f+\mathrm{id}_V)$ is trivial (contains only the zero vector).
One reason why you need $\mathrm{char}(F)\neq 2$ is that if $\mathrm{char}(F)=2$, then $1=-1$, so $f-\mathrm{id}_V = f+\mathrm{id}_V)$. And the sum of a subspace with itself cannot be a direct sum unless the subspace is trivial (which in this case would require $V$ to be trivial).
The second part is fairly easy: suppose $x$ is in the intersection. Then $(f+\mathrm{id}_V)(x) = \mathbf{0}$ and $(f-\mathrm{id}_V)(x) = \mathbf{0}$; the first equation tells you that $f(x)+x=\mathbf{0}$, the second that $f(x)-x=\mathbf{0}$. Now, since $2\neq 0$, you can conclude that $x=\mathbf{0}$.
So now you "just" need to show that every element of $V$ can be expressed as a sum of vectors in the appropriate place. Now, notice that so far we have not used the fact that $f\circ f=\mathrm{id}_V$, so presumably we are going to need to use it somehow. For one thing, what happens if you take an arbitrary vector $x$, and you look at $x-f(x)$? Then $f(x-f(x)) = f(x)-f(f(x)) = f(x)-x = -(x-f(x))$. So, does $x-f(x)$ lie in one of the two subspaces? What about $x+f(x)$? And, can you express $x$ as a linear combination of $x+f(x)$ and $x-f(x)$ (again, the fact that $2\neq 0$ is going to rear its ugly head...)