IBM tutorial's representation of a general unitary matrix $U(\theta,\phi,\lambda)$ can be derived as rotation of qubit on the Bloch sphere, in much the same way as the pdf reference has derived $R_{\hat{n}}(\alpha)$. But, these are two different ways of doing the same operation, requiring different user inputs. $R_{\hat{n}}(\alpha)$ considers rotation of qubit $|\psi\rangle$ about an arbitrary axis $\hat{n}$, whereas, $U(\theta,\phi,\lambda)$ is directly manipulating initial qubit state $|\psi\rangle$ to $|\psi '\rangle$.

The above figure shows qubit manipulation as rotation in a Bloch sphere representation in both ways. Either-
- The initial qubit state can be rotated about $Z$ axis by angle $\lambda$, then about $Y$ axis by angle $\theta$, and finally about $Z$ axis by angle $\phi$, to achieve $|\psi '\rangle$. In matrix form, this can be written as:
\begin{array}
\ U(\theta , \phi , \lambda) &= R_{Z}(\phi)R_{Y}(\theta)R_{Z}(\lambda) \\
&=
\begin{bmatrix}
e^{-i\phi/2} & 0\\
0 & e^{i\phi/2} \end{bmatrix}
\begin{bmatrix}
cos(\theta/2) & -sin(\theta/2)\\
sin(\theta/2) & cos(\theta/2)\end{bmatrix}
\begin{bmatrix}
e^{-i\lambda/2} & 0\\
0 & e^{i\lambda/2} \end{bmatrix}\\
&=
\begin{bmatrix}
cos(\theta/2)e^{-i\phi/2} & -sin(\theta/2)e^{-i\phi}\\
sin(\theta/2)e^{i\phi/2} & cos(\theta/2)e^{i\phi/2}
\end{bmatrix}
\begin{bmatrix}
e^{-i\lambda/2} & 0\\
0 & e^{i\lambda/2} \end{bmatrix}\\
&=e^{-i(\phi+\lambda)/2}
\begin{bmatrix}
cos(\theta/2) & -sin(\theta/2)e^{-i\lambda}\\
sin(\theta/2)e^{-i\phi} & cos(\theta/2)e^{i\{\phi+\lambda\}}
\end{bmatrix}\\
&=
\begin{bmatrix}
cos(\theta/2) & -sin(\theta/2)e^{-i\lambda}\\
sin(\theta/2)e^{-i\phi} & cos(\theta/2)e^{i\{\phi+\lambda\}}
\end{bmatrix}\mbox{, equal upto the global phase factor}
\end{array}
- Or, the operation of single qubit unitary gate can be visualized as rotation about arbitrary axis $\hat{n}$, i.e., first bringing $\hat{n}$ parallel to $|Z\rangle$, and then, rotating $|\psi\rangle$ by an angle $\alpha$ about $|Z\rangle$ axis, followed by bringing $\hat{n}$ back to its original position, as follows:
\begin{array}
\ R_{\hat{n}}(\alpha) &= f(\alpha, n_{\theta}, n_{\phi}) \\
&= R_Z(n_{\phi})R_Y(n_{\theta})R_Z(\alpha)R_Y(-n_{\theta})R_Z(-n_{\phi}) \\
&=
\begin{bmatrix}
cos(\alpha/2)-isin(\alpha/2)cos(n_{\theta}) & -isin(\alpha/2)sin(n_{\theta})e^{-i\phi}\\
-isin(\alpha/2)sin(n_{\theta})e^{i\phi} & cos(\alpha/2)+isin(\alpha/2)cos(n_{\theta}) \end{bmatrix}
\end{array}
I guess, the reason $U(\theta,\phi,\lambda)$ is an easier choice over $R_{\hat{n}}(\alpha)$ is because:
For a given initial and final qubit states, there is a unique magnitude of $(\theta, \phi, \lambda)$ representating unitary qubit gate, but, the same cannot be said about combination $(\alpha, \hat{n})$. This is because, any of the axis present on the perpendicular angle bisector plane, bisecting the angle between initial and final qubit Bloch vector, can represent that unitary gate. Off course, $\alpha$ will be decided on the choice of rotation axis chosen. $\alpha$ can be found out by noting the angle between projections of $|\psi\rangle$ and $|\psi '\rangle$ on plane normal to axis $\hat{n}$. (not shown in the figure).
Given a 2x2 unitary matrix, its much easier to find $(\theta, \phi, \lambda)$ than finding $(\alpha, n_{\theta}, n_{\phi})$. Though, one may argue that, finding $(\alpha, n_{x}, n_{y}, n_{z})$ is much easier. But, that's besides the point, and
Its far more initiative to think of directly rotating initial qubit to final state, then bringing a third vector into picture to do the job.
One may try finding relation between the two matrices which, in this analysis, essentially boils down to finding the relations between $(\theta, \phi, \lambda)$ and $(\alpha, n_{\theta}, n_{\phi})$.