Pretty self-explanatory I think; just wondering why neutral "manifests" from the windings and why it "outputs" (for lack of a better word) a negative-phase hot? I guess my perplexity comes from thinking that the [+hot/neutralwindings/-hot] are all the same wire so assuming that is correct what causes the changes?
The transformer doesn't manifest neutral: the ground rod does.
Or at least it should, according to electric codes. The "neutral" wiring is connected to the "safety ground" at exactly one point, and the safety ground is connected to this big copper rod stuck in the Earth. It's "neutral" because it's connected to Earth, and since you are probably also touching (perhaps indirectly) Earth, neutral won't shock you (absent any wiring faults, which is why there's a separate safety ground, which should be the same as neutral).
As a schematic, it looks like this:
Because the "neutral" is the center tap of the transformer, at any point the voltage at neutral will be exactly halfway between the voltages of the "hots". Thus, V1 is 120V, V2 is 120V, and if you add the two together you get 240V at V3. Or put another way, V1 = V2, both in terms of AC, or DC voltages at any instant.
The reason they are in opposite phase should be evident from the schematic: V1 has neutral on the "-" side, while V2 has it on the "+" side. As I've drawn the schematic, V1 and V2 are actually in phase, but if you flipped V2 around so that the "-" was on neutral, then they would be opposite phase. Because this is AC, which side is neutral and which is hot is largely irrelevant, except when you consider minor details like safety.
The transformer secondary consists of a really long wire. The 'hots' are connected to the ends of the wire, and the 'neutral' is connected to the middle. The wire is then wrapped into a coil wound around the transformer core. Current flowing through the primary side of the transformer creates an oscillating magnetic field in the transformer core. This field induces an electric field in the secondary. Since the whole winding experiences the same magnetic field (more or less), the voltage induced will be the same along the entire coil. If at one point in time 100V is induced along the entire coil and the 'neutral' tap is exactly in the middle, then you will see 50V between the 'neutral' and the 'hots'. Now, if you consider the 'neutral' to be 'zero potential', then one 'hot' will be at -50V and the other will be at +50V. You can just as easily consider one 'hot' to be at zero potential. In this case, one 'hot' will be at 0V, the 'neutral' at 50V, and the other 'hot' at 100V. The 'neutral' connection is really a matter of perspective. This relationship holds with any signal passing through the transformer. If you put an AC signal through and consider the 'neutral' to be 0V, then the two 'hots' will be inverted copies of each other. If you consider one of the 'hots' to be zero, then you will get two waveforms with the same phase out of the 'neutral' and the other 'hot', but the waveform coming out of the 'neutral' will have half of the amplitude.
Ignoring parasitic effects like resistance, leakage inductance, etc. one may regard an ideal transformer with a center-tapped secondary as being two transformers without center taps: one connects from the supply side to the outer two legs of the output; the other has one winding which connects the upper leg of the output to the "center", and the other of which connects the lower leg to the center.
An ideal transformer with a given turns ratio T will behave in such a fashion as to satisfy the equations:
V(sec) = -V(pri) * T I(sec) = I(pri) * T
Although real transformers only work with AC (as frequencies get lower, their behavior is increasingly dominated by parasitic effects, to the point that at DC they'd cease to be usable as transformers), an ideal transformer would work equally well at all frequencies from DC to daylight, and so it's sometimes useful to think in DC terms.
In our three-terminal center-tapped transformer, if the terminals are labeled H, L, and C, then, V(x) is the voltage at terminal x, and I(x) is the current flowing into terminal x, then
V(sec) = V(L)-V(C) V(pri) = V(H)-V(C) I(L) = I(sec) I(H) = I(pri) I(C) = -(I(L)+I(L))
Imagine that the transformer is driven with 1 volt between the two legs (with the "high" leg positive), and a 1 ohm resistor is connected between the positive leg and the center. The only thing which is connected to the lower leg is the supply. In this scenario, each winding will see half a volt across it, so the resistor across one of the windings will see half a volt as well, and will pass half an amp. Thus, half an amp will flow into the center terminal of the transformer. Because the two windings carry equal current, half the current will flow up, and half will flow down. The current (1/4 amp) which flows down will all recirculate through the supply (it has nowhere else to go), while the current which flows up (also 1/4 amp) will recirculate back through the load. Thus, the top side of the resistor is fed by 1/4 amp from the supply and 1/4 amp from the transformer.
Note that the fact that the center-tap is halfway between the two voltages is effectively forced by the fact that half of the current through the resistor is coming from the supply and half from the transformer's upper winding. Essentially, it causes the current through the resistor to be half of what it would be if it were connected directly across the supply, which in turn causes its voltage drop to be half of the supply voltage.
As noted, only ideal transformers work on DC; real ones do not. Thus, one would have to use AC (and accept its complications) to model the behavior of real transformers. Nonetheless, considering the behavior of an ideal transformer with DC should make it clear how a real transformer can feed current in or out of the neutral wire so as to keep its voltage halfway between those of the other wires.