Consider a graph like this:
Let's assume that there is no zero with a multiplicity greater than $3$. How can one tell the (least possible) degree? We know it's even, because the end behavior extends in the same direction. Adding up the multiplicities yields $6$, but how do we know it's not $8$, or $10$, or $276$ and is just translated in such a way that there are only $3$ x-intercepts?
Also, would counting turns and adding $1$ work in this case?
I wouldn't count turns. Think of the graph of $y=x^8.$
If a zero is simple, the graph crosses the $x$-axis like a line. In your graph, this happens at $x=9$. If a zero has even multiplicity, the graph won't cross the $x$-axis, but touch it and turn back. This happens at $x=4$. So there must be at least two more zeros. If a zero has odd multiplicity greater than one, the graph crosses the $x$-axis like a cubic. This happens at $x=-3.$ So that's at least three more zeros. So you polynomial has at least degree $6$.
If you graph $(x+3)^3(x-4)^2(x-9)$ it should look a lot like your graph.