If I understood well you are trying to deduce an expression for $\boldsymbol{A}$. I'll try to give it just by a retarded-time discussion without explicity passing from Lorentz transformations. You should consider a particle with charge $q_i$ moving along a trajectory $\boldsymbol{r}_i(t)$ with a velocity $\dot{\boldsymbol{r}}_i(\tau)\doteq\text{d}\boldsymbol{r}_i(\tau)/\text{d}\tau$ where $\tau$ will be the retarded time, defined at the end of the argumentation. Let's start from the equations of the Lorentz gauge
\begin{gather*}
\frac{1}{\mu}\nabla\cdot\boldsymbol{A}
+\varepsilon\frac{\partial\phi}{\partial t}=0
\\
\nabla^2\phi
-\mu\varepsilon\frac{\partial^2\phi}{\partial t^2}
=
-\frac{\rho}{\varepsilon}
\\
\nabla^2\boldsymbol{A}
-\mu\varepsilon\frac{\partial^2 \boldsymbol{A}}{\partial t^2}
=
-\mu\boldsymbol{j}
\end{gather*}
which have solutions
\begin{gather*}
\phi(\boldsymbol{r},t)
=
\frac{1}{4\pi\epsilon}
\int\limits_V
\frac{
\rho\left(\boldsymbol{r}^\prime,t
-\displaystyle{
\frac{|\boldsymbol{r}-\boldsymbol{r}^\prime|}{c}
}\right)
}{
|\boldsymbol{r}-\boldsymbol{r}^\prime|
}
\text{d}^3{\boldsymbol{r}^\prime}
\\
\boldsymbol{A}(\boldsymbol{r},t)
=
\frac{\mu}{4\pi}
\int\limits_V
\frac{
\boldsymbol{j}\left(\boldsymbol{r}^\prime,t
-\displaystyle{
\frac{|\boldsymbol{r}-\boldsymbol{r}^\prime|}{c}
}\right)
}{
|\boldsymbol{r}-\boldsymbol{r}^\prime|
}
\text{d}^3{\boldsymbol{r}^\prime}
\end{gather*}
but you know what $\rho,\boldsymbol{j}$ are for a moving particle
\begin{gather*}
\rho_i(\boldsymbol{r},t)
=
q_i\delta(\boldsymbol{r}-\boldsymbol{r}_i(t))
\\
\boldsymbol{j}_i(\boldsymbol{r},t)
=
q_i\boldsymbol{u}_i(t)\delta(\boldsymbol{r}-\boldsymbol{r}_i(t))
\\
\phi_i(\boldsymbol{r},t)
=
\frac{1}{4\pi\epsilon}
\int\limits_V
\frac{q_i\delta\left(\boldsymbol{r}^\prime-\boldsymbol{r}_i\left(t
-\displaystyle{\frac{|\boldsymbol{r}-\boldsymbol{r}^\prime|}{c}}\right)\right)
}{
|\boldsymbol{r}-\boldsymbol{r}^\prime|
}
\text{d}^3{\boldsymbol{r}^\prime}
\\
\boldsymbol{A}_i(\boldsymbol{r},t)
=
\frac{\mu}{4\pi}
\int\limits_V
\frac{q_i\dot{\boldsymbol{r}}_i\left(t-\displaystyle{\frac{|\boldsymbol{r}-\boldsymbol{r}^\prime|}{c}}\right)\delta\left(\boldsymbol{r}^\prime-\boldsymbol{r}_i\left(t
-\displaystyle{\frac{|\boldsymbol{r}-\boldsymbol{r}^\prime|}{c}}\right)\right)
}
{|\boldsymbol{r}-\boldsymbol{r}^\prime|
}
\text{d}^3{\boldsymbol{r}^\prime}
\end{gather*}
But you don't like very much this integral so you operate a substitution
\begin{gather*}
t^\prime
\doteq
t
-\displaystyle{\frac{|\boldsymbol{r}-\boldsymbol{r}^\prime|}{c}}
\\
\boldsymbol{d}_i(t)
\doteq
\boldsymbol{r}-\boldsymbol{r}_i(t)
\end{gather*}
such that the condition posed by the $\delta$ becomes $\boldsymbol{r}^\prime=\boldsymbol{r}-\boldsymbol{d}_i(t^\prime)$. So now you have
\begin{gather*}
\phi_i(\boldsymbol{r},t)
=
\frac{1}{4\pi\epsilon}
\int\limits_V
\frac{q_i\delta\left(\boldsymbol{r}^\prime-\boldsymbol{r}_i(t^\prime)\right)}{|\boldsymbol{d}_i(t^\prime)|}
\text{d}^3{\boldsymbol{r}^\prime}
\equiv
\frac{1}{4\pi\epsilon}
\int\limits_{t^{\prime\prime}}
\int\limits_V
\frac{q_i\delta\left(\boldsymbol{r}^\prime-\boldsymbol{r}_i(t^{\prime\prime})\right)}{|\boldsymbol{d}_i(t^{\prime\prime})|}
\delta(t^{\prime\prime}-t^\prime)
\text{d}^3{\boldsymbol{r}^\prime}\text{d}{t^{\prime\prime}}
\\
\boldsymbol{A}_i(\boldsymbol{r},t)
=
\frac{\mu}{4\pi}
\int\limits_V
\frac{q_i\dot{\boldsymbol{r}}_i(t^\prime)\delta\left(\boldsymbol{r}^\prime-\boldsymbol{r}_i(t^\prime)\right)}{|\boldsymbol{d}_i(t^\prime)|}
\text{d}^3{\boldsymbol{r}^\prime}
\equiv
\frac{\mu}{4\pi}
\int\limits_{t^{\prime\prime}}\int\limits_V
\frac{q_i\dot{\boldsymbol{r}}_i(t^{\prime\prime})\delta\left(\boldsymbol{r}^\prime-\boldsymbol{r}_i(t^{\prime\prime})\right)}{|\boldsymbol{d}_i(t^{\prime\prime})|}
\delta(t^{\prime\prime}-t^\prime)
\text{d}^3{\boldsymbol{r}^\prime}\text{d}{t^{\prime\prime}}
\\
\phi_i(\boldsymbol{r},t)
=
\frac{1}{4\pi\epsilon}
\int\limits_{t^{\prime\prime}}
\frac{q_i\delta\left(t^{\prime\prime}-t+\displaystyle{\frac{|\boldsymbol{r}-\boldsymbol{r}_i(t^{\prime\prime})|}{c}}\right)}{|\boldsymbol{d}_i(t^{\prime\prime})|}
\text{d}{t^{\prime\prime}}
\\
\boldsymbol{A}_i(\boldsymbol{r},t)
=
\frac{\mu}{4\pi}
\int\limits_{t^{\prime\prime}}
\frac{q_i\dot{\boldsymbol{r}}_i(t^{\prime\prime})\delta\left(t^{\prime\prime}-t+\displaystyle{\frac{|\boldsymbol{r}-\boldsymbol{r}_i(t^{\prime\prime})|}{c}}\right)}{|\boldsymbol{d}_i(t^{\prime\prime})|}
\text{d}{t^{\prime\prime}}
\end{gather*}
You see? Now the integral is on the time domain! We are near the end. Now a new variable substitution
\begin{equation*}
t^{\prime\prime\prime}
=
t^{\prime\prime}
-t
+\frac{|\boldsymbol{d}_i(t^{\prime\prime})|}{c}
\Longrightarrow
\text{d}{t^{\prime\prime\prime}}
=
\text{d}{t^{\prime\prime}}
+\frac{1}{c}\frac{\text{d}|\boldsymbol{d}_i(t^{\prime\prime})|}{\text{d}t^{\prime\prime}}\text{d}{t^{\prime\prime}}
\end{equation*}
but defining
\begin{equation*}
\boldsymbol{n}_i(t^{\prime\prime})
\doteq
\frac{\boldsymbol{d}_i(t^{\prime\prime})}{|\boldsymbol{d}_i(t^{\prime\prime})|}
\end{equation*}
You will see that
\begin{equation*}
{\text{d}|\boldsymbol{d}_i(t^{\prime\prime})|}{\text{d}t^{\prime\prime}}=-\boldsymbol{n}_i(t^{\prime\prime})\cdot\dot{\boldsymbol{r}}_i(t^{\prime\prime})
\end{equation*}
and define
\begin{equation*}
\kappa_i(t^{\prime\prime})
\doteq
1-\frac{1}{c}\boldsymbol{n}_i(t^{\prime\prime})\cdot\dot{\boldsymbol{r}}_i(t^{\prime\prime})
\end{equation*}
such that
\begin{gather*}
\phi_i(\boldsymbol{r},t)
=
\frac{1}{4\pi\epsilon}
\int\limits_{t^{\prime\prime\prime}}
\frac{q_i\delta\left(t^{\prime\prime\prime}\right)}{|\boldsymbol{d}_i(t^{\prime\prime})|\kappa_i(t^{\prime\prime})}
\text{d}{t^{\prime\prime\prime}}
\\
\boldsymbol{A}_i(\boldsymbol{r},t)
=
\frac{\mu}{4\pi}
\int\limits_{t^{\prime\prime\prime}}
\frac{q_i\dot{\boldsymbol{r}}_i(t^{\prime\prime})\delta\left(t^{\prime\prime\prime}\right)}{|\boldsymbol{d}_i(t^{\prime\prime})|\kappa_i(t^{\prime\prime})}
\text{d}{t^{\prime\prime\prime}}
\end{gather*}
and finally the last definition
\begin{gather*}
\tau
+\frac{|\boldsymbol{r}-\boldsymbol{r}_i(\tau)|}{c}
\doteq
t
\\
\phi_i(\boldsymbol{r},t)
=
\frac{1}{4\pi\epsilon}
\frac{q_i}{|\boldsymbol{r}-\boldsymbol{r}_i(\tau)|\kappa_i(\tau)}
\\
\boldsymbol{A}_i(\boldsymbol{r},t)
=
\frac{\mu}{4\pi}
\frac{q_i\dot{\boldsymbol{r}}_i(\tau)}{|\boldsymbol{r}-\boldsymbol{r}_i(\tau)|\kappa_i(\tau)}
\end{gather*}
That are the potential of Liénard-Wiechart. The expression of the electric field that you cited is just a consequence of Maxwell equations and so electromagnetic field can be obtained by the potentials, being careful with the gradient and the time derivative. Hope this helps
We know only scalar potential in both cases. My goal is to find vector potentials and to prove that
$$ \vec{E} = - \vec{\nabla} \phi - \frac{\partial \vec{A}}{\partial t} $$