(I assume that the OP wants to minimize is the *number* of turns rather than the total amount of absolute turning angles.)

If we restrict to moving only parallel to the axes, here is an elementary proof.

Assume there are $n$ rows and $m$ columns. Represent the traversing path by segments going from the center of one unit square to the center of the next and so on.
There are then three (overlapping) types of unit squares with path-segments, according to the diagram pictured (notice that end-of-path squares are considered TURNs):

[![Unit square types][1]][1]

**Trivial observation**: any row of the rectangle containing an X-square must contain at least $2$ TURN-squares. Likewise any column containing a Y-square must contain at least $2$ TURN-squares.

**Conclusion**: if each row contains at least one X-square, then there are at least $2n$ TURN-squares. Otherwise there is one row consisting entirely of Y-squares, in which case each column contains at least one Y-square, and then there are at least $2m$ TURN-squares.
Since $2$ of the $2n$ or $2m$ TURN-squares are end-of-path squares, there are at least $\min(2n,2m)-2$ true turns, as expected.

---------------------

**ADDENDUM** 7/8/2024. (As per comments, if I understood correctly.)

If any piecewise linear path in the plane is allowed (self-crossing, not grid-aligned, not contained in the minimal rectangle around the lattice points), then the above result doesn't hold. The picture shows an example for $m=n=4$, with only $5$ turns:

[![wild path][3]][3]


  [1]: https://i.sstatic.net/nDKWd1PN.png
  [2]: https://i.sstatic.net/2QiA03M6.jpg
  [3]: https://i.sstatic.net/lTPbVG9F.jpg