Firstly, you might be interested in paper Elementary gates for quantum computation explaining how complex gates can be decomposed to simpler ones. This would allow you understand how the matrix $U_j$ is decomposed.
Before we proceeed further, we have to define gate $U1$ used on IBM Q computer:
$$
U1(\lambda)=
\begin{pmatrix}
1 & 0 \\
0 & e^{i\lambda}
\end{pmatrix}
$$
While the authors present matrix $U_j$, the implementation of acutal gate in the figure is its controlled version. Therefore, is it a little bit different.
The controlled version of the first matrix in the decomposition correspond to the first gate in the picture. On IBM Q, it is controlled $U1$ gate with $\lambda = c-a$.
The second gate is composed of global phase gate with angle $a$ and $U1$ gate with angle $b-a$. The global phase $e^a$ is simply factored out of the matrix which leaves you with $e^a\text{diag}(1;e^{b-a})$. So controlled version of this gate is implemented with $U1$ having $\lambda = a$ applied on the control qubit and no gate (or rather identity gate $I$) applied on the target qubit. This is the controlled global phase gate. Then it is followed by controlled $U1$ with angle $b-a$.
The last matrix in the decomposition is controlled $U1$ gate described by 4x4 matrix
$$
\begin{pmatrix}I & O \\ O & U1\end{pmatrix},
$$
where $I$ is identity matrix 2x2 and $O$ is zero matrix 2x2.
Since the whole $U_j$ has to be controlled, the last gate is controlled controlled $U1$ (or double controlled $U1$) implemented with last three gates in the figure. However, this part seems a little bit strange in the original paper and I think it is wrong. According to the lemma 6.1 in the paper linked above, it is possible to construct double controlled gate with CNOT gates and gates $V$ and $V^\dagger$ such that $V^2 = U$. In this case $U = U1(a)$, therefore $V = U1(a/2)$ and $V^\dagger = U1(-a/2)$. The reason is that $U1$ is a kind of a rotation, so $V^2=VV=U1(a/2)U1(a/2)=U1(a)$ and inverse to $U1(a)$ is $U1(-a)$. You can check all this by direct calculation.
Also note that exponent in the last matrix is wrong. It should be $d-c+a-b$. You can check this by calculating right side of the equation. As the matrices are presented in the paper, it is impossible to arrive back to $U_j$ on the left side.
With the decomposition provided in the question, lemma 6.1 I mentioned and correction to the exponent, the correct code for controlled version of $U_j$
should be
\\first matrix
u1(c-a) x,y;
\\second matrix
u1(a) x;
cu1(b-a) x,z;
\\third matrix (with lemma 6.1)
cu1((d-c+a-b)/2) y,z;
cx x,y;
cu1(-(d-c+a-b)/2) y,z;
cx x,y;
cu1((d-c+a-b)/2) x,z;
Finally, I would list all mistakes in the paper:
- the exponent $x$ should be $d-c+a-b$, not $d+c-a-b$
- there should not be Toffoli gates but CNOTs on qubits $x$ and $y$
- there should be sign minus before parameter of $U1$ on seventh row
- no Toffoli gate should be in the circuit diagram and $U(x)$ should be double controlled gate
You can also find some additional information about the paper in this thread.