You may understand the velocity field $\mathbf{v}:\mathbb{R}^3\times\mathbb{R}\to\mathbb{R}^3$ and the acceleration field $\mathbf{a}:\mathbb{R}^3\times\mathbb{R}\to\mathbb{R}^3$ in the following way.
$\mathbf{v}=\mathbf{v}(\mathbf{x},t)$ is called a velocity field if it reproduces the velocity of the particle that appears at $\mathbf{x}$ at the moment $t$. That is, if $\mathbf{y}=\mathbf{y}(t)$ tracks the motion of a particle, then
$$
\frac{\rm d}{{\rm d}t}\mathbf{y}(t)=\mathbf{v}(\mathbf{y}(t),t).
$$
In this last formulation, the left-hand-side is obviously the velocity of this particle, while the right-hand-side is how $\mathbf{v}$ reproduces this velocity.
Similarly, $\mathbf{a}=\mathbf{a}(\mathbf{x},t)$ is called an acceleration field if it reproduces the acceleration of the particle that appears at $\mathbf{x}$ at the moment $t$. Still let $\mathbf{y}=\mathbf{y}(t)$ track the motion of a particle, then
$$
\frac{\rm d^2}{{\rm d}t^2}\mathbf{y}(t)=\mathbf{a}(\mathbf{y}(t),t).
$$
So much about the definitions. Now let us figure out the relation between $\mathbf{a}$ and $\mathbf{v}$. We will use the above two definitions repeatedly.
\begin{align}
\mathbf{a}(\mathbf{y}(t),t)&=\frac{\rm d^2}{{\rm d}t^2}\mathbf{y}(t)=\frac{\rm d}{{\rm d}t}\left(\frac{\rm d}{{\rm d}t}\mathbf{y}(t)\right)\\
&=\frac{\rm d}{{\rm d}t}\mathbf{v}(\mathbf{y}(t),t)\\
&=\sum_{j=1}^3\frac{\partial\mathbf{v}}{\partial x_j}(\mathbf{y}(t),t)\frac{{\rm d}y_j}{{\rm d}t}(t)+\frac{\partial\mathbf{v}}{\partial t}(\mathbf{y}(t),t)\\
&=\left(\frac{\rm d}{{\rm d}t}\mathbf{y}(t)\cdot\nabla\right)\mathbf{v}(\mathbf{y}(t),t)+\frac{\partial\mathbf{v}}{\partial t}(\mathbf{y}(t),t)\\
&=\left(\mathbf{v}(\mathbf{y}(t),t)\cdot\nabla\right)\mathbf{v}(\mathbf{y}(t),t)+\frac{\partial\mathbf{v}}{\partial t}(\mathbf{y}(t),t).
\end{align}
Finally, note that $\mathbf{y}(t)$ could be arbitrarily chosen, and that for any chosen $\mathbf{y}(t)$, the above identity must hold. Therefore, we conclude that
$$
\mathbf{a}(\mathbf{x},t)=\left(\mathbf{v}(\mathbf{x},t)\cdot\nabla\right)\mathbf{v}(\mathbf{x},t)+\frac{\partial\mathbf{v}}{\partial t}(\mathbf{x},t),
$$
or simply put it as
$$
\mathbf{a}=\left(\frac{\partial}{\partial t}+\mathbf{v}\cdot\nabla\right)\mathbf{v}.
$$
Obviously, this is nothing but the material derivative. This is not a surprise at all, because the governing Euler equations (or Navier-Stokes equations) are derived based on the second law of Newton, namely $m\mathbf{a}=\mathbf{F}$.
Back to the question. Following your notation $\mathbf{a}=\left(a_x,a_y,a_z\right)$ and $\mathbf{v}=\left(v_x,v_y,v_z\right)$, the formulae $\mathbf{a}=\left(\partial_t+\mathbf{v}\cdot\nabla\right)\mathbf{v}$ mean that
\begin{align}
a_x&=\left(\frac{\partial}{\partial t}+v_x\frac{\partial}{\partial x}+v_y\frac{\partial}{\partial y}+v_z\frac{\partial}{\partial z}\right)v_x,\\
a_y&=\left(\frac{\partial}{\partial t}+v_x\frac{\partial}{\partial x}+v_y\frac{\partial}{\partial y}+v_z\frac{\partial}{\partial z}\right)v_y,\\
a_z&=\left(\frac{\partial}{\partial t}+v_x\frac{\partial}{\partial x}+v_y\frac{\partial}{\partial y}+v_z\frac{\partial}{\partial z}\right)v_z.
\end{align}
The rest would be straightforward calculations.
Last but not least, as commented in above, even if $\mathbf{v}$ is time-independent, $\mathbf{a}$ could still be non-vanishing. This is, mathematically, due to the contribution of the advection term $\left(\mathbf{v}\cdot\nabla\right)\mathbf{v}$. An intuitive interpretation could be that $\mathbf{v}$ provides an overall picture of the macro scope of a flow. It could reproduce the motion of each particle therein, but this does not mean that it equals the motion of that particle.
