First of all, let's prove the theorem, then talk about what it means. The first direction, let $E$ be idempotent, that is $E^2 = E$. This, at least to me, seems at like a kinda opaque condition, it's not very geometric. Well, let's make it geometric.
First off, idempotence implies that anything inside the range is fixed by $E$, that is $E|_{R(E)} = id$. To see this, suppose $y \in R(E)$, write $y = E(x)$, then we have $E(y) = E(E(x)) = E(x) = y$
Observe that $N(E) \cap R(E) = 0$, indeed, suppose $y$ lies in both, we have $y = E(y)$ by $y \in R(E)$, and $E(y) = 0$, by $y \in N(E)$!
So far, no linear algebra. Now it comes in.
Okay, so the kernel and range only touch at 0, and by rank-nullity the sum of their dimensions is $n$, so if we choose a basis $e_1, \ldots e_k$ for the kernel, and a basis $f_1, \ldots f_{n-k}$ for the range, then $e_1 \ldots e_k, f_1, \ldots f_{n-k}$form a basis for $\mathbb{C}^n$ (this really has nothing to do with $\mathbb{C}$, you could replace it with $\mathbb{R}$, or $\mathbb{Q}$, etc.), and our map $E$ now looks like $\sum \lambda_i e_i + \sum \lambda_j f_j \mapsto \sum \lambda_j f_j$i.e. we forget our coordinates in the kernel. As a matrix, this looks like 0's everywhere except for 1s along the diagonal for the coordinates $f_j$. For example, if $k = n-k = 1$, we have $\begin{bmatrix} 0 & 0 \\ 0 & 1 \end{bmatrix}$or if say $k = 1, n-k = 2$, we have $\begin{bmatrix} 0 & 0 & 0 \\ 0 & 1 & 0 \\ 0 & 0 & 1 \end{bmatrix}$The reason why it's called projection is that we take a vector $v$ and send it to its shadow in $R(E)$, for example if $k = n-k = 1$ as above, we take a vector $(a,b)$ and send it to $(a,0)$. Pictorially, the vector forms the hypotenuse of the right triangle with vertices $\{(0,0), (a,0), (a,b)\}$, we're "squishing" the hypotenuse down to one leg with vertices $\{(0,0), (a,0)\}$.
In the other direction, about $L$ and $M$ complementary subspaces, we can form an idempotent by writing down a matrix just like those above, namely take a basis $e_1, \ldots e_k$ for $L$ and $f_1, \ldots f_{n-k}$ for $M$, and write down the same matrix.
So the takeaway is this: the condition $E^2 = E$ isn't that mysterious, it just amounts to choosing a subspace $M$, to "cast shadows onto", and a subspace $N$, to determine "where the sun sits in the sky", i.e. which complementary subspace casts no shadow.