Weierstrass equations are probably a good choice. You can try
$$y^2 = x^3 - 3x +2 t^{12},$$ for example.

Here the singular fibers are when $t$ is a $24$th root of unity $\zeta$, and the double point is provided by $x = \zeta^{12}$ and $y=0$.

Aren't the points where $d\pi$ is not injective the same as the singular points?

To really describe this as a smooth surface over $\mathbb P^1$, first note that we can get a surface in $\mathbb P^2 \times \mathbb A^1$ by homogenizing, i.e. taking $(x:y:z)$ to be the coordinates of $\mathbb P^2$ and $t$ the coordinate of $\mathbb A^1$, look at the vanishing locus of the equation $$y^2 z= x^3 - 3xz^2 +2 t^{12}z^3$$ and then another chart is provided by changing variables to $\tilde{x} = x/ t^4$, $\tilde{y} = y/t^6$, $\tilde{z} = z$, $\tilde{t} =1/t$, where after the change of variables we are looking at the vanishing locus of the equation 
$$\tilde{y}^2 \tilde{z} = \tilde{x}^3 - 3 \tilde{t}^8 \tilde{x} \tilde{z}^2 +2 \tilde{z}^3$$ in $\mathbb P^2 \times \mathbb A^1$ with coordinates $(\tilde{x}:\tilde{y}:\tilde{z}),\tilde{t}$.

For me a hyper-Kähler structure is just a nowhere vanishing holomorphic 2-form. Such a 2-form is provided here by $$\frac{dx}{y} \wedge dt = \frac{d (t^4 \tilde{x} )}{t^6\tilde{y} } \wedge dt = \frac{t^4}{t^6} \frac{d\tilde{x}}{\tilde{y}} \wedge dt = - \frac{d\tilde{x}}{\tilde{y}}\wedge d \tilde{t}.$$