TL;DR
You have rediscovered the Bloch sphere! :-)
Interesting special cases
Before we prove that the map defined by equation $(1)$ is the representation of single-qubit quantum states as points on the Bloch sphere, it is instructive to consider a few special cases. Let us denote the map with $\phi: \mathcal{H} \to \mathbb{R}^3$ where $\mathcal{H}$ is the Hilbert space of a qubit. We calculate that
$$
\phi(|0\rangle) = \phi\left(\begin{pmatrix}1 \\ 0\end{pmatrix}\right) = \begin{pmatrix}
2 \cdot (1 \cdot 0 + 0 \cdot 0) \\
2 \cdot (-0 \cdot 0 + 1 \cdot 0) \\
1^2 + 0^2 - 0^2 - 0^2
\end{pmatrix} = \begin{pmatrix} 0\\0\\1 \end{pmatrix} \\
\phi(|1\rangle) = \phi\left(\begin{pmatrix}0 \\ 1\end{pmatrix}\right) = \begin{pmatrix}
2 \cdot (0 \cdot 1 + 0 \cdot 0) \\
2 \cdot (-0 \cdot 1 + 0 \cdot 0) \\
0^2 + 0^2 - 1^2 - 0^2
\end{pmatrix} = \begin{pmatrix} 0\\0\\-1 \end{pmatrix} \\
\phi(|+\rangle) = \phi\left(\begin{pmatrix}\frac{1}{\sqrt{2}} \\ \frac{1}{\sqrt{2}}\end{pmatrix}\right) = \begin{pmatrix}
2 \cdot \left(\frac{1}{\sqrt{2}} \cdot \frac{1}{\sqrt{2}} + 0 \cdot 0\right) \\
2 \cdot \left(-0 \cdot \frac{1}{\sqrt{2}} + \frac{1}{\sqrt{2}} \cdot 0\right) \\
\left(\frac{1}{\sqrt{2}}\right)^2 + 0^2 - \left(\frac{1}{\sqrt{2}}\right)^2 - 0^2
\end{pmatrix} = \begin{pmatrix} 1\\0\\0 \end{pmatrix} \\
\phi(|-\rangle) = \phi\left(\begin{pmatrix}\frac{1}{\sqrt{2}} \\ -\frac{1}{\sqrt{2}}\end{pmatrix}\right) = \begin{pmatrix}
2 \cdot \left(\frac{1}{\sqrt{2}} \cdot \left(-\frac{1}{\sqrt{2}}\right) + 0 \cdot 0\right) \\
2 \cdot \left(-0 \cdot \left(-\frac{1}{\sqrt{2}}\right) + \frac{1}{\sqrt{2}} \cdot 0\right) \\
\left(\frac{1}{\sqrt{2}}\right)^2 + 0^2 - \left(-\frac{1}{\sqrt{2}}\right)^2 - 0^2
\end{pmatrix} = \begin{pmatrix} -1\\0\\0 \end{pmatrix} \\
\phi(|{+i}\rangle) = \phi\left(\begin{pmatrix}\frac{1}{\sqrt{2}} \\ \frac{i}{\sqrt{2}}\end{pmatrix}\right) = \begin{pmatrix}
2 \cdot \left(\frac{1}{\sqrt{2}} \cdot 0 + 0 \cdot \frac{1}{\sqrt{2}}\right) \\
2 \cdot \left(-0 \cdot 0 + \frac{1}{\sqrt{2}} \cdot \frac{1}{\sqrt{2}}\right) \\
\left(\frac{1}{\sqrt{2}}\right)^2 + 0^2 - 0^2 - \left(\frac{1}{\sqrt{2}}\right)^2
\end{pmatrix} = \begin{pmatrix}0\\1\\0 \end{pmatrix} \\
\phi(|{-i}\rangle) = \phi\left(\begin{pmatrix}\frac{1}{\sqrt{2}} \\ -\frac{i}{\sqrt{2}}\end{pmatrix}\right) = \begin{pmatrix}
2 \cdot \left(\frac{1}{\sqrt{2}} \cdot 0 + 0 \cdot \left(-\frac{1}{\sqrt{2}}\right)\right) \\
2 \cdot \left(-0 \cdot 0 + \frac{1}{\sqrt{2}} \cdot \left(-\frac{1}{\sqrt{2}}\right)\right) \\
\left(\frac{1}{\sqrt{2}}\right)^2 + 0^2 - 0^2 - \left(-\frac{1}{\sqrt{2}}\right)^2
\end{pmatrix} = \begin{pmatrix}0\\-1\\0 \end{pmatrix}.
$$
Though not a proof, the above equalities are very suggestive.
Moreover, these equalities explain why the three components of $\langle\psi|\vec{\sigma}|\psi\rangle$ correspond to the three components of $\phi(|\psi\rangle)$. For example, $|+\rangle$ and $|-\rangle$ are eigenvectors of $\sigma_x$, so $\phi$ maps them to vectors that are fixed by the corresponding 3D rotation of $\mathbb{R}^3$.
Proof in the general case
You have almost obtained the proof in the general case when you switched from representing an arbitrary quantum state as $|\psi\rangle=(a + bi, c + di)^T$ to representing it as $|\psi\rangle=(\cos\frac{\theta}{2}, e^{i\phi}\sin\frac{\theta}{2})$ and calculated that
$$
\begin{align}
\phi(|\psi\rangle) &= \phi\left(\begin{pmatrix}\cos\frac{\theta}{2} \\ e^{i\phi}\sin\frac{\theta}{2}\end{pmatrix}\right) \\
& =\begin{pmatrix}
2 \cos\frac{\theta}{2} \sin\frac{\theta}{2} \cos\phi \\
2 \cos\frac{\theta}{2} \sin\frac{\theta}{2} \sin\phi\\
\cos^2\frac{\theta}{2} - \sin^2\frac{\theta}{2} \cos^2\phi - \sin^2\frac{\theta}{2} \sin^2\phi
\end{pmatrix} \\
&= \begin{pmatrix}
\sin\theta\cos\phi \\
\sin\theta\sin\phi \\
\cos\theta
\end{pmatrix}.
\end{align}
$$
Now the only missing step is to notice that the last result is the point on the unit sphere in $\mathbb{R}^3$ with azimutal angle $\phi$ and polar angle $\frac{\theta}{2}$, exactly as on the Bloch sphere.
Relationship to the Lie group homomorphism $SU(2)\to SO(3)$
A key property of the Bloch sphere is that for any unitary $U \in SU(2)$ and any state $|\psi\rangle \in \mathcal{H}$, the 3-vector $\phi(U|\psi\rangle)$ can be obtained from the 3-vector $\phi(|\psi\rangle)$ by rotating the latter using $\Phi(U)\in SO(3)$ where $\Phi: SU(2) \to SO(3)$ denotes the Lie group homomorphism that maps single-qubit unitaries in $SU(2)$ to 3D rotations in $SO(3)$.
This fact can be expressed as
$$
\phi(U|\psi\rangle) = \Phi(U)\phi(|\psi\rangle)\tag2
$$
or by saying that the following diagram
$$
\begin{array}{ccccc}
&\mathcal{H} & \xrightarrow{\phi} & S^2 & \\
U&\Bigg\downarrow & & \Bigg\downarrow & \Phi(U) \\
&\mathcal{H} & \xrightarrow[\phi]{} & S^2 & \\
\end{array}
$$
commutes for every $U\in SU(2)$. The proof of $(2)$ is a straightforward, but somewhat lengthy calculation.