How to interpret adfuller test results?

The typical way to reject the null hypothesis would be that your t-test result -1.84 is less than all critical values (1%, 5%, 10%) which in this case, it's not less than your critical values.

what you stated in your question is correct. Once you applied the Adfuller test over your OLS regression residue, you were checking whether your residue had any heterocedasticity, in another words, if your residue was stationary.

Since your adfuller p-value is lower than a certain specified alpha (i.e.: 5%), then you may reject the null hypothesis (Ho), because the probability of getting a p-value as low as that by mere luck (random chance) is very unlikely.

Once the Ho is rejected, the alternative hypothesis (Ha) can be accepted, which in this case would be: the residue series is stationary.

Here is the hypothesis relation for you:

Ho: the series is not stationary, it presents heterocedasticity. In another words, your residue depends on itself (i.e.: yt depends on yt-1, yt-1 depends on yt-2 ..., and so on)

Ha: the series is stationary (That is normally what we desire in regression analysis). Nothing more is needed to be done.

Null hypothesis: Non Stationarity exists in the series.

Alternative Hypothesis: Stationarity exists in the series

Data: (-1.8481210964862593, 0.35684591783869046, 0, 1954, {'10%': -2.5675580437891359, 
'1%': -3.4337010293693235, '5%': -2.863020285222162}, 21029.870846458849)

Lets break data one by one.

First data point: -1.8481210964862593: Critical value of the data in your case

Second data point: 0.35684591783869046: Probability that null hypothesis will not be rejected(p-value)

Third data point: 0: Number of lags used in regression to determine t-statistic. So there are no auto correlations going back to '0' periods here.

Forth data point: 1954: Number of observations used in the analysis.

Fifth data point: {'10%': -2.5675580437891359, '1%': -3.4337010293693235, '5%': -2.863020285222162}: T values corresponding to adfuller test.

Since critical value -1.8>-2.5,-3.4,-2.8 (t-values at 1%,5%and 10% confidence intervals), null hypothesis cannot be rejected. So there is non stationarity in your data

Also p-value of 0.35>0.05(if we take 5% significance level or 95% confidence interval), null hypothesis cannot be rejected.

Hence data is non stationary (that means it has relation with time)