The reason that the second diagram you drew cannot represent what is happening is that it will not satisfy Newton's 2nd law for all parcels of fluid in the tank. Imagine that you had a tank like the one shown in the diagrams and, rather than accelerating it, you just tilt it at an angle so that base is no longer horizontal. Basically, what you ave done is change the direction of gravity relative to the sides of the container. Would you expect the water surface to remain parallel to the base of the container, or would you expect it to be horizontal again (but tilted relative to the base). What you have done in the acceleration experiments is to add a pseudo-gravitational force component in the direction opposite to the acceleration. So now, the effective gravity is no longer pointing in the vertical direction. Thus, the surface of the fluid must readjust to again be perpendicular to the new effective gravitational direction (which is not vertical).
If you do a force balance on a small parcel of fluid within the system having sides dx, dy, and dz, the force balance in the y (vertical) direction reduces to:$$\frac{\partial p}{\partial y}=-\rho g$$The force balance in the x (horizontal) direction reduces to:$$\frac{\partial p}{\partial x}=-\rho a$$where a is the acceleration. The variation of pressure with position is given by: $$dp=\frac{\partial p}{\partial x}dx+\frac{\partial p}{\partial y}dy=-\rho adx-\rho g dy$$ It follows from this that the surfaces of constant pressure are given by:$$\frac{dy}{dx}=-\frac{a}{g}$$ The free surface is a contour of constant pressure.