How do multiple mixing stages work in a spectrum analyzer?
A true multiplicative mixer produces components at both the sum and difference of the frequencies (real mixers will usually produce components at other frequencies too but we can ignore that for now)). Generally filters will be arranged so that only one of those components passes through to the next stage.

Frequencies (in this context) are always positive, so difference here means absoloute value of the difference.

In this case the filters are not shown, but from the frequencies labeled on the diagram they are clearly always taking the difference and filtering out the sum. 

So writing this out as equations.

$$f_\mathrm{if1} = |f_\mathrm{sig}-f_\mathrm{lo1}|$$

$$f_\mathrm{if2} = |f_\mathrm{if1}-f_\mathrm{lo2}|$$

$$f_\mathrm{if3} = |f_\mathrm{if2}-f_\mathrm{lo3}|$$

If we combined these equations together and tried to solve them we would end up with 8 possible soloutions for \$f_\mathrm{sig}\$ in terms of \$f_\mathrm{if3}\$. Which of those 8 soloutions is correct? well we need to go back to our diagram.

$$f_\mathrm{sig} &lt; f_\mathrm{lo1} \rightarrow f_\mathrm{if1} = f_\mathrm{lo1} - f_\mathrm{sig}$$

$$f_\mathrm{if1} &gt; f_\mathrm{lo2} \rightarrow f_\mathrm{if2} = f_\mathrm{if1} - f_\mathrm{lo2}$$

$$f_\mathrm{if2} &gt; F_\mathrm{lo3} \rightarrow f_\mathrm{if3} = f_\mathrm{if2} - f_\mathrm{lo3}$$

Combing these equations.

$$ f_\mathrm{if3} = ((f_\mathrm{lo1} - f_\mathrm{sig}) - f_\mathrm{lo2})- f_\mathrm{lo3} $$

Getting rid of the brackets.

$$ f_\mathrm{if3} = f_\mathrm{lo1} - f_\mathrm{sig} - f_\mathrm{lo2}- f_\mathrm{lo3} $$

Rearrange to make \$f_\mathrm{sig}\$ the subject.

$$ f_\mathrm{sig} = f_\mathrm{lo1} - f_\mathrm{lo2}- f_\mathrm{lo3} - f_\mathrm{if3}$$

Which is equvilent to

$$ f_\mathrm{sig} = f_\mathrm{lo1} - (f_\mathrm{lo2} + f_\mathrm{lo3} + f_\mathrm{if3})$$

P.S curiously the diagram labels the first Lo as 5.1 to 8.7 GHz but to get the frequencies shown in the rest of the diagram would require a first LO of 8.7225 GHz.