How should species density be calculated for a clumped distribution?
To me, there are two issues that are mixed up here (if I understand you correctly). First, do you want to estimate the mean and variance for a statistical population (i.e. to characterize a larger population by independent samples), or do you want to calculate the actual density for a particular area, where you have counted all occurences in that entire area (but maybe divided the area into subareas out of convenience when counting)? This is not clear from your question.

In the second case, your second option of pooling counts and areas is suitable. However, then you have only calculated the actual density in that particular area (ignoring issues of detectability of the organism when counting), and you cannot draw any inferences about a larger statistical population.

If your aim is to draw inferences for a larger statistical population, a start is to calculate the mean and standard deviation (sd) of your sample. In that case, I assume that your samples are chosen randomly and independently from a larger statistical population. Your first option is then the right approach. However, since your samples have different sizes, you might want to attach more weight to larger samples, since they can be assumed to better describe the population average than smaller samples. This is called a weighted mean.

Generally, the weighted arithmetic mean is defined as: 

$$\bar{x}_w = \dfrac{\sum_{i=1}^nw_ix_i}{\sum_{i=1}^nw_i}$$

where $w_i$ are the weights for each sample ($x_i$) and $n$ are the number of samples. Using this formula, you will arrive at exactly the same value for the weighted mean as you calculated in your second attempt (0.458), if you use the areas as weights. 

The weighted standard deviation is a bit more problematic, in the sense that there doesn't exist one single standard way of calculating this. However, a commonly used formula is:

$$ \sigma_{w_1} = \sqrt{ \frac{ \sum_{i=1}^n w_i (x_i - \bar{x}_w)^2 }{ \frac{(M-1)}{M} \sum_{i=1}^n w_i } },$$

where $M$ are the number of nonzero weights. Other definitions of the weighted standard deviation can be found at Wikipedia: weighted sample variance.

Another version is defined as (called "reliability weights" on the wiki page):

$$ \sigma_{w_2} = \sqrt{ \frac{ \sum_{i=1}^n w_i (x_i - \bar{x}_w)^2 }{  \sum_{i=1}^n w_i - \frac{\sum_{i=1}^n w_i^2}{\sum_{i=1}^n w_i}}},$$

If I haven't made a mistake, these will give the standard deviations 1.15 and 1.22, using your data.

As for the biological interpretation, all calculations of average densities indicate a clumped distribution, since the coefficient of variation (CV) is larger than 1 (CV = sd/mean), but more so if you use a weighted mean. The CV using the arithmetric mean is 1.5, while it is 2.5-2.65 for the weighted mean, which is reasonable since you are giving more weight to the large area sample with a zero count. However, I should also note that you should be cautious about using a weighted mean when you have a strongly clumped distribution, since your run the risk that e.g. a large sample plot land on an area with no occurences, which might bias your density estimate. Generally, when you have a clumped distribution, you need to sample more intensively to get a good estimate of average density, and many small samples is often better than a few large ones.