I know that acetic acid is a stronger acid than phenol because in acetic acid, the charge is delocalized on two oxygen atoms which is better than 1 oxygen and 3 carbon.
But in case of para-nitrophenol, $\ce{NO2}$ reduces the negative charge in the ring, thereby stabilizing the intermediate formed.
So does this mean that the stabilization caused by the presence of $\ce{NO2}$ is less than the delocalization caused by 2 oxygen atoms?
Kindly correct me if I am going in the wrong direction.
The difference in acidity between acetic acid and phenol can be attributed to the electron-withdrawing properties of a carbonyl function versus a phenyl ring. In both cases, the deprotonated oxygen can be depicted in a resonance structure as forming a double bond to the carbon in question and the charge being delocalised within the ring.
It should immediately strike anybody that a phenyl ring primarily provides a resonance body (i.e. a greater volume into which the charge can be delocalised) while a carbonyl group is also actively withdrawing (both inductively and mesomericly). This is reflected when comparing the $\mathrm{p}K_\mathrm{a}$ values of $\ce{CH3CO-CH3}$ (acetone) and $\ce{Ph-CH3}$ (toluene) — $16.5$ logarithmic units in $\mathrm{DMSO}$ according to the Evans table[1] or approximately $24$ logarithmic units in water ($\mathrm{p}K_\mathrm{a}$ of acetone in water taken from the Brodwell data).[2]
Adding a nitro group to the phenyl ring reduces this difference drastically to $-6.1$ logarithmic units (i.e. para-nitrotoluene is more acidic than acetic acid) — in $\ce{DMSO}$. One should always remember that acidity differences in one solvent can be drastically different in another solvent. For example, the difference between $\ce{MsOH}$ and $\ce{TfOH}$ is $11.4$ logarithmic units in water but only $1.3$ logarthmic units in $\ce{DMSO}$ (that is not a typo). One should also note the difference between acetic acid in water ($\mathrm{p}K_\mathrm{a}\approx4.76$) and $\ce{DMSO}$ ($\mathrm{p}K_\mathrm{a}\approx12.3$).
$$\textbf{Table 1: }\text{Acidity constants of various compounds in various media}\\\text{All data taken from the Evans data unless noted otherwise in the text.}$$ \begin{array}{lrc}\hline \text{compound} & \mathrm{p}K_\mathrm{a}(\ce{H2O}) & \mathrm{p}K_\mathrm{a}(\ce{DMSO})\\ \hline \ce{CH3CO-CH3} & 19.3\phantom{000} & 26.5\\ \ce{Ph-CH3} & 31.5\phantom{000} & 30.6\\ \ce{$p${-}NO2-C6H4-CH3} & & 20.4\\ \ce{MsOH} & -2.6\phantom{000} & \phantom{0}1.6 \\ \ce{TsOH} & -14\phantom{.0000} & \phantom{0}0.3 \\ \ce{AcOH} & 4.76\phantom{00} & 12.3\\ \hline \end{array}
Given that a $\mathrm{p}K_\mathrm{a}$ value of para-nitrotoluene has not been reported in water (although it’s $\mathrm{p}K_\mathrm{b}$ value has been),[3] we are not in any position to draw definite conclusions. However, water being the solvent it is, I would still assume the electron-withdrawing ability of a nitro group which relies solely on mesomeric effects to be weaker than the effect of a directly neighbouring carbonyl group. Remember that water is excellent at stabilising charges; $\ce{DMSO}$ not so much.
References:
[1]: The Evans $\mathrm{p}K_\mathrm{a}$ table
[2]: The Brodwell $\mathrm{p}K_\mathrm{a}$ data
[3]: J. C. D. Brand, J. Chem. Soc. 1950, 997. DOI: 10.1039/JR9500000997.
So, this is a pretty complicated topic. Your reasoning is actually pretty good, but you are missing some key data. Comparing a phenol to a carboxylic acid is somewhat hard.
The first, simplest, and least satisfying answer is that the $\ce{NO2}$ is farther away from the acidic hydrogen in 4-nitrophenol than the oxygen is from the acidic hydrogen in acetic acid. Not good enough? Let's dig further.
We might decide to expand upon your argument that because 4-nitrophenol has one more oxygen in it than acetic acid, there is more potential for electron withdrawal, and hence 4-nitrophenol should be stronger. Solid reasoning, but consider our first, less satisfying answer. Both of these effects might be combined, and we could reason that the distance effect is stronger than the increase in electron withdrawing power. Decent reasoning, but still unsatisfying.
Enter the Hammett Equation:
$$\log \frac{K}{K_0} = \sigma \rho$$
Fully explaining this takes up one or two chapters of a graduate level physical chemistry book. Quick summary, $K_0$ refers to a reference reaction, which has hydrogen as its substituent. $\sigma$ is the electron withdrawing(positive) or electron donating (negative) value of a substituent replacing the hydrogen. $\rho$ refers to a parameter that is descriptive of the reaction itself. The parameters are empirically determined relative to the acidity of substituted benzoic acids. When benzene rings are not involved, sometimes an equivalent called the Taft parameter is used. Run 6 reactions (5 plus reference), draw a line, get your rho value for any reaction.
However, the reference reaction is the dissociation of benzoic acid. $\rho$ is different for the dissociation of acetic acid and for nitrophenol. So, even though nitro is a much stronger withdrawing group, its influence is mitigated by a reaction much less subject to such effects. The Wikipedia article is not terrible on this matter.
On acetic acid, the carbon in the carboxylic acid group is highly electron-starved: that carbon is bonded to oxygen atoms 3 times! Moreover, that carbon only has one methyl group from which to "take" electrons in order to fill the charge imbalance. This means that the only easy electrons the acidic oxygen on the carboxylic acid group can access is the one on the hydrogen. This makes the O-H bond more ionic in character, and so easy to dissociate in water.
On phenol, the carbon in the hydroxyl group is not electron-starved. Since it is part of an electron-rich aromatic ring, it is a "channel" through which the oxygen atom can access electrons - this lucky oxygen atom can access 6 electrons that are delocalized over 6 carbon atoms in order to satisfy its hunger for negative charges. This means that the O-H bond in question is more covalent in character, and more difficult to dissociate in solution.
