I can't get my head around the following term:
$(\frac{\partial}{\partial t} + v_o\nabla)^2 p(t,x)$
I know what $\frac{\partial}{\partial t}p(t,x)$ is supposed to mean and I assume that $v_o\nabla p(t,x)$ is simply a product of $v_o$ with $p(t,x)$ derivated in space. But how can I solve the bracket?
First of all you have to consider that what you have inside the bracket is an operator.
This means that you cannot compute the normal square of the two objects like $a^2 + b^2 + 2ab$, but you will have to write $ab$ and $ba$ differently, because they will be two different objects.
To see what I mean, let's expand the square:
$$\left(\frac{\partial}{\partial t} + v_0\nabla\right)^2 = \frac{\partial^2}{\partial t^2} + v_0^2 \nabla^2 + v_0\frac{\partial}{\partial t}\nabla + v_0 \nabla\frac{\partial}{\partial t}$$
In which my only assumption has been that $v_0$ is a scalar, that is, a number.
So, in another way, you can collect it and get
$$\frac{\partial^2}{\partial t^2} + v_0^2 \nabla^2 + v_0\left(\frac{\partial}{\partial t}\nabla + \nabla\frac{\partial}{\partial t}\right)$$
Hence, when acting of the function $p(t,x)$ you will get
$$\frac{\partial^2 p(x, t)}{\partial t^2} + v_0^2 \nabla^2p(t, x) + v_0\left(\frac{\partial}{\partial t}\nabla + \nabla\frac{\partial}{\partial t}\right)p(t, x)$$
And the last fracked will obviously mean
$$v_0\left(\frac{\partial}{\partial t}\nabla p(t, x) + \nabla\frac{\partial}{\partial t}p(t, x)\right)$$
Then it clearly depends upon the function $p$ you have. Under certain conditions, the two operators will commute that is
$$\frac{\partial}{\partial t}\nabla = \nabla \frac{\partial}{\partial t}$$
And in this case you will be the normal double product, but it's not always guaranteed.