Can we use the work-energy theorem to find pressure function a liquid in this case?

Let this be the situation:

For pressure difference between two vertically displaced points:

Let's solve in the frame of the container, therefore we take pseudo force.

For reference:$P_1+rho gh_1+frac{1}{2}rho v_1^2=P_2+rho gh_2+frac{1}{2}rho v_2^2$

Applying conditions and simplifying,$$P_1-P_2=rho (g+a_y)(h_2-h_1)$$

Here $h_2-h_1$ is the separation between any two points vertically

Now for pressure difference between two points displaced horizontally :

We can rotate our reference frame by $90^circ$ so now our "gravity " along the horizontal axis will be $a_x$

Appliying conditions and simplyfiyng:$$P_1-P_2=rho (a_x)(x_2-x_1)$$

Here $x_2-x_1$ is the separation between two horizontal points.

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Yes, it is possible but we will have to adjust our acceleration term accordingly. It is usually $g$ since we don't use Bernoulli's equation often on accelerating bodies

Also note $g,a_x,a_y$ are all negative as we take them as a pseudo force.

Corrections are welcomed.