Kitz is perfectly right with her explanation, but I'll add one more element in... the symbol rate, which helps tie the "sum of the bit-loading" to the final throughput.
In a DMT system, all those parallel modems are actually transmitting 4,000
symbols per second. Every single modem (for its individual tone) transmits one symbol every 0.25 msecs.
What is a symbol?
In analogue terms, it is simply a 0.25ms-long waveform, with a specific phase and amplitude. The DSL modem's job is to transmit or detect this waveform.
In digital terms, that analogue waveform gets translated to/from a number of bits - the exact number being determined by the "bit loading" of the tone. One analogue symbol will map to 1-15 bits, depending on the bit-loading of that tone.
For tones where there is a lot of signal and little noise (high SNR), the receiving modem can detect finer-grained differences between the phase & amplitude values of the waveforms ... so is capable of transferrining more bits, and the "bit loading" can be higher. When SNR is low, receiver has a harder job distinguishing differences, so the bit loading must be kept low.
The best way I have seen to picture the translation between an analogue symbol and the number of bits is through two-dimensional graphs known as constellations. These give a translation between an N-bit number and the corresponding phase-angle+modulation.
The picture in this section of wikipedia gives a simple example of a 4-bit constellation (don't worry about the text, just look at the animated image).
https://en.wikipedia.org/wiki/Quadrature_amplitude_modulation#Quantized_QAMHere's a picture showing a few different constellations for bit-loads of 1-5. Look at figure 4:
https://www.osapublishing.org/oe/fulltext.cfm?uri=oe-19-26-B486&id=224908It actually shows results applicable to fibre, rather than DSL, but the principle is the same.
Figure 6 this page gives a few more examples:
http://140.98.202.196/xpls/icp.jsp?arnumber=6479217&pgName=figuresSo ... the answer to your quandary about bit-loading is that you needed to calculate the sum of bits, as you did, which tells you how many bits are transferred per symbol (ie every 0.25ms). Multiplying this by 4,000 gives you the number of bits per second.
That still doesn't help you entirely, because other line-coding details get in the way, such as trellis encoding, and FEC, and framing. But it gets you closer.