Both ammonium and sodium bicarbonate solutions are basic, as the anion $\ce{HCO3-}$ is a stronger base than is the strength of the acid $\ce{NH4+}$, not even talking about $\ce{Na+}$.
Carbonic acid could form a mildly acidic salt in sense of forming acidic solution with a base weaker than $\ce{HCO3-}$, e.g. with pyridine or aniline. But their water solutions would be highly hydrolyzed to the respective base and $\ce{H2CO3(aq)/CO2(aq)}$, as such bases need very acidic pH. Carbon dioxide would be gradually escaping as gas.
$$\ce{CO2(aq) + H2O(l) <=> H2CO3(aq)} \tag{1}$$
$$\ce{H2CO3(aq) + H2O(l) <=> HCO3-(aq) + H3O+(aq)}\tag{2}$$
$$\ce{C5H5N(aq) + H3O(aq) <=> C5H5NH+(aq) + H2O(l)}\tag{3}$$
Very formally, the salt formula can be written as $\ce{(C5H5NH)HCO3}$, as it would in large extent hydrolyze to the left side of reactions (2) , (3) to pyridyne and carbon dioxide.
An acidic salt – aside of pH of its solution – means a salt where not all acidic hydrogens of a multiprotic acid are neutralized. Like
- $\ce{KHSO4}$ (very acidic solution)
- $\ce{KH2PO4}$ (acidic solution)
- $\ce{NaHCO3}$ (mildly basic solution)
- $\ce{Na2HPO4}$ (basic solution)
As the anion of such partially neutralized acid is both an acid ($\ce{HA- <=> H+ + A^2-}$) and a base ($\ce{HA- + H+ <=> H2A}$), pH of its solution depends on which of them is stronger.
Similarly, basic salts are combination of a salt and a hydroxide, like the bleaching powder is a complex combination of hydrated calcium chloride, hypochlorite and hydroxide.
Wikipedia:
Commercial calcium hypochlorite consists of anhydrous $\ce{Ca(ClO)}$2, dibasic calcium hypochlorite $\ce{Ca3(ClO)2(OH)4}$ (also written as $\ce{Ca(ClO)2·2Ca(OH)2}$), and dibasic calcium chloride $\ce{Ca3Cl2(OH)4}$ (also written as $\ce{CaCl2·2Ca(OH)2}$).
Or there is $\ce{CuCO3 . Cu(OH)2}$ in the mineral malachite.
How can someone figure out which of the bases is stronger between $\ce{HCO3-}$ and Pyridine. And if we have the assertion that pyridinium bicarbonate will be acidic in nature, then is the reason that pyridine is a weaker base than $\ce{HCO3-}$ hence pyridinium ion will be stronger acid than $\ce{HCO3-}$?
The values of acidity/basicity constanst can be found tabulated. The most available ways is getting them from Wikipedia pages for respective compounds, or various chemistry dedicated sited have listed them for many compounds.
If there is an equilibrium
$$\ce{HA(aq) <=> H+(aq) + A-(aq)}$$
or
$$\ce{B(aq) + H+(aq) <=> BH+(aq)},$$
then acidity of $\ce{HA}$ or $\ce{BH+}$ and basicity of $\ce{A-}$ or $\ce{B}$ are linked by the equation
$$K_\mathrm{a} \cdot K_\mathrm{b} = K_\mathrm{w},$$
resp. (at $\ce{25 ^{\circ}C}$)
$$\mathrm{p}K_\mathrm{a} + \mathrm{p}K_\mathrm{b} = 14$$.
If there are two bases, then the stronger one forms by neutralization a weaker acid. Pyridinium with $\mathrm{p}K_\mathrm{a} = 5.23$ is a stronger acid than CO2(aq) with $\mathrm{p}K_\mathrm{a1}=6.35$, therefore pyridine is a weaker base than $\ce{HCO3-}$.