Why do we interpret the first term of the Fokker-Planck equation as drift? With the derivation of the Fokker-Planck equation we get:
$$\frac{\partial}{\partial t}P(x,t)=-\frac{\partial}{\partial x}(A(x,t)P(x,t))+\frac{1}{2}\frac{\partial^2}{\partial x^2}(B(x,t)P(x,t))$$
We then interpret the first term as the drift with $A(x,t)$ as drift velocity and the second term as diffusion with $B(x,t)$ as $2D$ with $D$ as the diffusion coefficient. The second term looks similar to the diffusion equation, so I understand this part of the interpretation. What is the reason for the drift interpretation?
 A: That term doesn't just appear - it comes from introducing a drift term in the first step of the derivation from your link. 
$${\displaystyle dX_{t}=\mu (X_{t},t)\,dt+\sigma (X_{t},t)\,dW_{t}}$$
The $\mu$ is a mean drift velocity which leads to a $dx = v*dt$ relationship in that equation, while the second term models diffusion. So to answer your question, we aren't interpreting it once we get that final result; we know it is a drift term from our initial assumptions in writing that first step down. 
A: As already stated in alex1stef2's answer, $A$ comes from the literal drift term in the SDE of the process. 
But, let's look at this another way and assume you were only given the FP-equation without knowledge of its derivation.
If you start the dynamics with an initial deterministic distribution $P(z,0)=\delta(z-x0)$ and set $B\equiv 0$, you'll find that
$ P(z,t) = \delta(z-x(t)) $ where $x(t)$ is the solution of the ordinary differential equation
$$ \dot{x} = A(x) $$ i.e. the deterministic equation of motion with initial value $x_0$.
In the absence of the diffusion term, the distribution stays deterministic and moves according to the drift prescribed by $A$.
See also: Gardiner, Stochastic Methods for a derivation and discussion.

Addendum: Plugging in $\delta(z-x(t))$ on either side of the Fokker-Planck-equation yields
1.) $$ \partial_t \delta(z-x(t)) = -\partial_z\delta(z-x(t))\frac{dx}{dt} = -\partial_z\delta(z-x(t))A(x(t)) $$
2.) $$-\partial_z\left[A(z)\delta(z-x(t))\right] = -\partial_z\left[A(x(t))\delta(z-x(t))\right] = -A(x(t))\partial_z\delta(z-x(t)) $$
The first equality in 2.) is due to the delta-function. Check this if you are unsure. 
Note that I have renamed $x$ to $z$ from the orginal answer to avoid ambiguity. 
