why is the current -1A at the negative terminal, I thought current should have the same value of 1A throughout this circuit?
Current is a charge flow rate and since flow is directional, it is a vector quantity.
In a circuit, the direction is restricted to be along the conductor, so we only need that one dimension, but it can still be positive if it flows in the direction we designated or negative if it flows in the opposite direction.
Now when we designate the direction as out of the battery, the current will be positive on the positive terminal, meaning the current flows out of the battery, and negative on the negative terminal, meaning it returns into the battery there.
The magnitude is the same, but we are concerned also about the direction in this case.
Secondly how does it make sense to talk about the current through the whole surface of the battery, I don't understand what it means by "the surface of the battery including both terminals".
It is not actually true that the current should have the same value throughout the circuit. If charge is being accumulated somewhere, the current in and out of that somewhere will be different. 531
For example in a static discharge the current flows from the place depleted of electrons to another place that accumulated electrons in an open circuit until the charges equalize.
Expressed mathematically
$$ \frac{\mathrm{d}q}{\mathrm{d}t} = I_{in} - I_{out} $$
that is rate of charge accumulation is current in minus current out. Using current as vector allows us to simplify this to just
$$ \frac{\mathrm{d}q}{\mathrm{d}t} = -\sum I_{out} $$
by taking the flow in as negative as your professor did there.
Now with the talk about surface you are switching from the simplified view of circuits as elements connected with one-dimensional wires to full three-dimensional analysis. It would make sense to start talking about current density, which is the current through some area – think cross-section of a conductor – divided by that area.
That allows us to write an even more general form of the above equation
$$ \frac{\mathrm{d}q}{\mathrm{d}t} = -\oint_A j\,\mathrm{d}A $$
That is the rate of charge accumulation inside some volume is the current flow through its surface times normal vector of that surface.
The integral is +1A through the positive terminal, -1A through the negative terminal and 0 through the rest of the surface as that is insulated, resulting in overall zero and showing that the battery as a whole stays electrically neutral during operation. 531