Why does steric hindrance cause a d8 complex to have a tetrahedral geometry rather than a square planar geometry?
For a given number of ligands surrounding a central atom there is always an arrangement in which the distances between the ligands is maximised and which typically also results in the maximum possible angles between then. You can deduce them logically if you start at a single ligand (which has no steric clashes whatsoever).

A second ligand should approach the same centre from exactly the opposite side leading to a linear arrangement and a bond angle of $180^\circ$.

A third ligand should approach the same centre perpendicular to the existing bond axis. Upon its approach, the two ligands already present will ease away from the approaching one. The final arrangement is trigonal planar with bond angles of $120^\circ$.

A fourth ligand should approach the same centre in a perpendicular manner again; this time relative to the plane of the three existing ligands. Again, they will ease away to make more room for the incoming. The final arrangement is a tetrahedron with bond angles of $109.5^\circ$.

Going from a tetrahedron to a square-planar arrangement means squashing the ligands together i.e. reducing the space between them. Thus, if there are no electronic arguments favouring a different arrangement any tetracoordinated entity should assume a tetrahedral environment to minimise steric strain.