To get a really good idea of what is going on, we need a space "large enough" to have some interesting projections. So let's look at $\Bbb R^3$. Since the image of any linear transformation (including a projection) is a subspace, it follows that any projection is either:

1) the 0-map (not interesting)

2) an isomorphism (also not interesting)

3) projection onto a plane (this is interesting)

4) projection onto a line (also interesting).

To over-simplify things a bit, suppose we choose bases such that $E_1$ is the (orthogonal) projection onto the $xy$-plane, and $E_2$ is the (orthogonal) projection onto the $z$-axis. This means that $\mathfrak{M}_1 = \Bbb R^2 \times \{0\}$, and $\mathfrak{M_2} =\{(0,0)\} \times \Bbb R$. One thing you should be aware of is that:

$\mathfrak{N}_j = \text{ker }E_j$ for $j = 1,2$.

So let's look explicitly at what $E_1 + E_2$ is, explicitly, in matrix form:

$E_1 + E_2 = \begin{bmatrix}1&0&0\\0&1&0\\0&0&0 \end{bmatrix} + \begin{bmatrix}0&0&0\\0&0&0\\0&0&1 \end{bmatrix} = \begin{bmatrix}1&0&0\\0&1&0\\0&0&1 \end{bmatrix} = I$.

Since the identity transformation is (trivially) a projection (of $\Bbb R^3$ along the 0-vector), we should expect that $E_1E_2 = E_2E_1 = 0$ (which I urge you to verify).

Hopefully it is clear that $\mathfrak{M}_1 \oplus \mathfrak{M}_2 = \Bbb R^3$, and that $\mathfrak{N}_1 \cap \mathfrak{N}_2 = \{(0,0,0)\}$

In a case like this, we say $E_1$ and $E_2$ are complementary projections (because they "split the identity cleanly").

To see an illustration of part (ii) of the theorem choose $E_2$ as above (orthogonal projection onto the $z$-axis), and choose $E_1 = I$.

To see an illustration of part (iii) take $E_1$ to be the projection onto the $xy$-plane, and $E_2$ to be the projection onto the $y$-axis. I urge you to explicitly calculate the image and kernel of $E$ for both parts (ii) and (iii).