I figure that there are better answers possible, but this might do for a stater:
Assuming that your $f$ and $g$ are minimimal (i.e. there is no $u$, with $\deg(u)>0$ s.t $u^4$ divides $f$ and $u^6$ divides $g$), then your Weierstrass equation defines a K3 surface if and only if $\deg(f)\leq 8$ and $\deg(g)\leq 12$ and at least one of $\deg(f)\geq 5, \deg(g)\geq 7$ holds. E.g. $y^2=x^3+z^{12}-1$ defines a smooth K3 surface.
This example also contradicts your claim about the fiber type. The above equation defines a smooth surface and has 12 II fibers. Smoothness of the Weierstrass equation implies only that each singular fiber is of type $I_1$ or $II$, not necessary of type $I_1$.
One of the things you seem to look for is the monodromy action on $H^1(E,\mathbb{Z})$ for $E$ a smooth fiber. You can find this in many places e.g. Barth-Hulek-Peters-Van de Ven Chapter V. Sec 7-13.
The rest of the problem can be reformulated in terms of the Picard-Fuchs differential equation for a family of elliptic curves. There is a book of Stiller studying this, but I have to admit that I never read that book. Texts explaining mirror symmetry to mathematicians contain often a description how to calculate PF equation. (E.g., the book of Cox and Katz.)