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 $$\frac{dx}{y} \wedge dt = \frac{d (t^4 \tilde{x} )}{t^6\tilde{y} } \wedge dt = \frac{t^4}{t^6} \frac{d\tilde{x}}{\tilde{y}} \wedge dt = - \frac{d\tilde{x}}{\tilde{y}}\wedge d \tilde{t}.$$

Let me explain how to make Jason Starr's construction explicit, modulo constructing a splitting. I'll construct two linearly independent complex-valued (but not holomorphic) vector fields, which give four linearly independent real vector fields.

Let $v_1$ be the (meromorphic) vector field on the elliptic surface, which is contained in the elliptic curve fibers, and which is inverse to the 1-form $dx/y$. In other words, if $u$ is any coordinate on the elliptic curve fiber, $v_1$ is given by $ y (dx/du)^{-1} \partial/\partial u$. A classical argument shows $dx/y$ is well-defined and non-degenerate everywhere, even on the points where $y$ or $dx/du$ vanishes or has a pole (since the zeroes and poles cancel).

Let $v_2$ be any smooth lift of the vector field $\partial/\partial t$ on $\mathbb A^1$, i.e. a vector field whose projection onto the $t$ line has size $1$ at each point. Such a $v_2$ exists by a partition of unity argument, but I don't know a formula for it. (Away from the singular fibers, a lift can be constructed using the natural flat torus structure on the fibers, flowing one into the next in the unique locally-affine zero-preserving way, but this vector field has poles at the singular fibers.)

Then $v_1$ has a pole of order $2$ at $t = \infty$, and no other poles or zeroes away from the $24$ singular points, while $v_2$ has a zero of order $2$ at $\infty$. Thus $v_2 / t^2 = v_2 \overline{t}^2/ |t|^4$ is well-defined and nonvanishing in a neighborhood of $t=\infty$, so $v_2 \overline{t}^2 / (1+ |t|^2)^2$ is well-defined and nonvanishing in a neighborhood of $t= \infty$.

Then $$ \frac{v_1}{ (1 + |t|^2)^2} + t^2 v_2 $$ and $$ \frac{v_1 \overline{t}^2 }{ (1 + |t|^2)^2} + i v_2$$

are well-defined complex-valued vector fields, and linearly independent at each nonsingular point:

Since $v_1$ and $v_2$ are well-defined and linearly independent away from $\infty$, these two linear combinations are linearly independent away from the vanishing locus of the determinant $\frac{i}{ (1+|t|^2)^2} + \frac{ t^2 \overline{t}^2}{(1+|t|^2)^2} = \frac{ |t|^4 +i}{ (1+ |t|^2)^2} $, which is never vanishing.)

At $t=\infty$, $ \frac{v_1 \overline{t}^2 }{ (1 + |t|^2)^2}$ and $t^2 v_2 $ are well-defined and linearly-independent, and $ \frac{v_1}{ (1 + |t|^2)^2} +$ and $i v_2$ vanish, so the two vector fields are well-defined at $t=\infty$ as well.

Then the real and imaginary parts of the two vector fields above are linearly independent and thus give a framing of the K3 surface minus the $24$ singular points. However, they can't be made explicit without an explicit choice of lift $v_2$.