If you mean that you don't understand how to go from the first line to the second, all they've done is to insert the identity operator in the form ##1=\sum_k|k\rangle\langle k|##. If you mean that you don't understand how to go from the second row to the third, then you need to learn about the relationship between linear operators and matrices.
Let U and V be vector spaces. Let ##T:U\to V## be linear. Let ##A=(u_1,\dots,u_n)## and ##B=(v_1,\dots,v_m)## be ordered bases for U and V respectively. The matrix [T] of the linear operator T with respect to the pair (A,B) of ordered bases is defined by
$$[T]_{ij}=(Tu_j)_i.$$ The right-hand side is interpreted as "the ith component of the vector ##Tu_j## in the ordered basis B".
You should find it very easy to verify that if B is an orthonormal ordered basis, we have ##[T]_{ij}=\langle v_i,Tu_j\rangle##.
The reason for the definition of [T] can be seen by doing a simple calculation. Suppose that Tx=y. I won't write any summation sigmas, since we can remember to do a sum over each index that appears twice.
$$\begin{align}y &=y_i v_i\\
Tx &= T(x_j u_j)=x_jT(u_j)=x_j(Tu_j)_i v_i.\end{align}$$ Since the v_i are linearly independent, this implies that ##y_i=(Tu_j)_i x_j##. This can be interpreted as a matrix equation [y]=[T][x], if we define [y] and [x] in the obvious ways, and [T] as above. (Recall that the definition of matrix multiplication is ##(AB)_{ij}=A_{ik}B_{kj}##).
When U=V, it's convenient to choose A=B, and we can talk about the matrix of a linear operator with respect to an ordered basis, instead of with respect to a pair of ordered bases.
Notations like [T] are typically only used in explanations like this. I think most books would use T both for the linear operator and the corresponding matrix with respect to a pair of ordered bases.
Edit: It would be a good exercise for you to prove that if ##T:U\to V## and ##S:V\to W## are linear, then ##[S\circ T]=[T]##. This result is the main reason why matrix multiplication is defined the way it is.
