I wonder if you are mixing up manifolds, metric and coordinate systems.
Mathematicians have a precise definition of a manifold, but for our purposes it's just something that sets the dimensionality. We're normally only interested in the manifold $\mathbb{R}^4$, though I guess other manifolds would creep in if we were dealing with spacetimes with non-trivial topologies.
The metric defines distance on the manifold. It allows us to measure the distance along a curve between any two points. The metric is coordinate independent, so it's wrong to say:
It is the metric that is doing the separation of space and time
The metric will have a signature that specifies how many timelike and spacelike dimensions there are, and we are normally only interested in metrics with the signature (3,1), or (1,3) depending on your sign convention. These have one timelike and three spacelike dimensions, but how we split spacetime up into timelike and spacelike parts is done by choosing a coordinate system.
And finally, the coordinate system is how we choose to measure out space and time. The most intuitive example is to use three orthogonal spatial axes $x$, $y$ and $z$ and a time axis $t$, but this is far from the only choice. For example in describing the spacetime geometry round a black hole we may use the obvious simple coordinates (usually as polar coordinates), but we might choose Gullstrand-Painleve, Eddington-Finkelstein or Kruskal-Szekeres coordinates and probably lots of others that I can't think of. These split up time and space in different ways.
The point of all this is that if you take any curve in spacetime you can divide it up into a series of infinitesimal straight lines $\mathbf{ds}$ then calculate the length of $\mathbf{ds}$ using:
$$ ds^2 = g_{\alpha\beta}dx^\alpha dx^\beta $$
where $g_{\alpha\beta}$ is the metric tensor and the $dx^\alpha$ are our four coordinates. For example in flat spacetime $g$ is the Minkowski metric and we can write the equation as:
$$ ds^2 = -c^2dt^2 + dx^2 + dy^2 + dz^2 $$
The values of $g$ and the $dx$s will depend on what coordinate system we have chosen, however the key point is that no matter what coordinates we choose the value of $ds^2$ we end up with is always the same.
So regardless of what coordinates we use we can always tell if a curve is timelike just by writing down an expression for $ds^2$ anywhere along its length and then seeing if $ds^2$ is positive, negative or zero. And we can use whatever coordinate system is most convenient because the answer doesn't depend on what coordinates we choose.
So we can tell if any random curve can be the world line of a massive object just by showing that $ds^2$ is everwhere negative along the curve. Similarly we can show the curve is the trajectory of a massless particle by showing that $ds^2$ is everywhere zero on the curve.
However, although we can choose whatever bizarre coordinates we want, we cannot do the calculation unless we have a metric. That's because without a metric the curve is just a set of spacetime points with no concept of distance between points on the curve.
I appear to have ranted on a bit: I wasn't sure exactly what you were asking so I just wrote down anything that seemed relevant. If I've misunderstood what you are asking please say so ina comment and I'll edit the answer accordingly.