I know this not that well posed of a question so please bear with me. Suppose we have a $n$-dimensional psuedo-Riemannian manifold $(M,g)$. We have that there are $n^2$ functions that make up $g=[g_{ij}]$. However not all of them can be chosen arbitrarily, since $g_{ij}=g_{ji}$. So that leaves $\frac{n(n+1)}{2}$ functions remaining. Now my question for an arbitrary $g$, how many of those functions are truly "independent", and determine $g$.
The reason I ask is because I was looking at the case where $n=2$. One can always find isothermal coordinates for $(M,g)$ in which $g_{ij}=\Lambda\delta_{ij}$, where $\Lambda$ is a smooth function and $\delta_{ij}$ is the kronecker delta. So in the case of $n=2$ we have that $(M,g)$ has 1 degree of freedom.
I have heard in passing that in general that there are $\frac{n(n-1)}{2}$ degrees of freedom. If someone could help me either word this better, give me resources or offer a heuristic argument as to why that is true it would be great.
The answer is quite trivial. In $n$ dimensions you are free to change the coordinates by providing $n$ independent functions. So the actual number of the degrees of freedom for the metric is $$n(n+1) / 2 - n = n(n-1) /2.$$ The answer, which only depends on the general rank discussion, is therefore the same for the pseudo-metric as well.
You have implicitly mentioned this fact already for $n=2$ when you said one can find isothermal coordinates. This is essentially nothing more than diagonalization procedure. What can be surprising is that for $n > 2$ the diagonalization can fail for some metrics. Even though from linear algebra we know that the metric can be diagonalized at every point, over bigger portions of the manifold this need not be the case.
In the general theory of relativity this phenomenon manifests as a frame dragging caused by the $g_{\phi t}$ term of the metric.