How does standardising a secondary standard solution make it suitable as a standard solution?
When talking about standards, there's always the dilemma: How is a standard standardized? Well, you compare it to another standard you already have, and after you consider all the sources of errors, you have a new standard.

In case of the primary standards, you can always promptly trust the label's purity information, as well as your balance's and other measuring equipment (like glassware) accuracy, and at a much lower degree the molar masses, and so on. Having that concentration "certified", anything you calculate using that as a reference will be "certified" as well. However...

If you're going to follow strict analytical rules, than you won't blindly trust those, which is why you cshould calculate the error (or many of the related statistical measures) by considering the accuracy of, like I said, the balance, the glassware, temperature effect on volume (density), etc. 

One common way to do this, called the standardized uncertainty, is take the square root of the sum of the squares of each instrument/measure uncertainty divided by it's quantity. You should define a minimal formula which describes the property you want to account the error for, and go for there. Let's take, for example, the concentration of 100 mL of a 0.01 M solution of potassium hydrogen phthalate (KHP), used as a primary standard to standardize sodium hydroxide.

The concentration of the solution is $[KHP]=\frac{n}{vol}$, but we obviously didn't measure molês, so we write

$$[KHP]=\frac{\frac{m}{MM}}{vol}$$

where $m$ is the mass weighed on the balance, and $MM$ the molar mass. Now that we know what influence in your concentration, we have to use some numbers. Let's consider that:


Weighed mass: 0.20422 g, on a balance with uncertainty of ± 0.001 g (1 mg)
Volume measured: 100 mL, in a volumetric flask with uncertainty of ± 0.1 mL


This means that the uncertainty of the prepared solution is

$$\frac{u_{[KHP]}}{0.01} = \sqrt{\left(\frac{0.001}{0.20422}\right)^2 + \left(\frac{0.1}{100}\right)^2} $$

$$u_{[KHP]} \approx 5\times10^{-5}$$

Now note that I didn't take into account the purity of the salt (to make things simpler), and the error on the molar mass (we could assume it's much lower than the others). And what does all this have to do with your question? 

My point is, even standards have a variations in them, and when you standardize something, like a HCl or NaOH solution, you could take that into account. The concentration of NaOH, in this case, will have a relative uncertainty higher than the concentration of KHP, because all of the uncertainty of the KHP preparation must be taken into account, as well as the titration variation. The outcome is that a standard based on  a standard of another standard (...) will eventually have a very high uncertainty, which is why we try to sitck to primary and secondary standards. 

PS.: The uncertainties of the equipments/glassware are usually available in the manuals or even engraved on them. This values are usually the calibration precision, and should be transformed into uncertainty using some statistical inference, which could be that the variation follows a retangular distribution, resulting in

$$u = \frac{precision}{\sqrt3}$$

Disclaimer: There's a lot more to this than what I summarized above, starting with being able to indentify what should and shouldn't be taken into account, not to mention replicates and a lot of statistics. I don't recommend blindly following this formulae to calculate uncertainties.