For many large cardinals, it is indeed true that the order we usually use (consistency strength over ZFC) coincides with this ordering: The least inaccessible is strictly weaker than the least Mahlo, which is strictly weaker than the least weakly compact, for example. But this ordering does not behave well for all large cardinals.

Magidor showed that it is consistent (relative to the existence of supercompact cardinals) that the smallest strongly compact cardinal is also the smallest measurable cardinal. He also showed that it is consistent that the smallest strongly compact cardinal is the smallest supercompact. Now: The smallest supercompact is limit of measurables, so we have a problem if we use your ordering, as the position of "strong compactness" would depend on the model we are considering, which is not what one would like, namely, that the ordering depends only on the axioms and the theories they imply.

Another example is that we may have a model with Wodin cardinals, and no strong cardinals, and models with many strong cardinals but no Woodins.

The ordering in terms of consistency strength is more subtle. It coincides (as far as we know) with a natural ordering of certain inner models. Independently of whether the least strongly compact is the least measurable or not, if there is a strongly compact, there are inner models with many measurable cardinals (and much more). This shows that in a clear sense, "strong compactness" is a stronger assumption.

The issue with Woodin and strong cardinals is similar, but there is an additional subtlety, namely, (simplifying matters a tiny bit) to be a Woodin cardinal is a "local" assumption, we just need a large enough initial segment of $V$ to verify the Woodiness of a given cardinal. However, to be a strong cardinal is a "global" assumption", that requires that we check the whole universe. Something similar occurs with supercompactness (global) and other assumptions, like the existence of rank into rank embeddings(local). What is true is that, independently of size, if there is a Woodin cardinal, then there are inner models with many strong cardinals, so the assumption that there is a Woodin is strictly stronger.

Moreover, we are more interested in "large cardinal assumptions" than in "large cardinals" per se, and for many assumptions, largeness in terms of cardinality is not a good measure. For example: The existence of $0^\sharp$ is a large cardinal assumption. It is significantly stronger than the existence of a weakly compact cardinal (for example, if $0^\sharp$ exists, then there is a proper class of weakly compact cardinals in $L$). However, $0^\sharp$ is a countable object, while weakly compacts are not.

The follow up question is what large cardinal assumption is the largest (in whichever appropriate ordering we end up using). At the moment, such an assumption does not seem to exist. We have a template for strong large cardinal axioms, namely describing properties of elementary embeddings. We know not everything is possible: Kunen showed there are no Reinhardt cardinals, i.e., no embeddings $j:V\to V$. This gives us an upper bound on what is possible. We have been considering assumptions that in a sense look very close to this boundary (rank into rank embeddings, for example), but it may well be that closeness is a feature of the resolution of our current telescopes. Years ago, it was expected that supercompactness was an assumption just slightly above the existence of strong cardinals. Nowadays we see a huge chasm in between.

For many large cardinals, it is indeed true that the order we usually use (consistency strength over ZFC) coincides with the ordering you suggest: The least inaccessible is strictly weaker than the least Mahlo, which is strictly weaker than the least weakly compact, for example. But this ordering does not behave well for all large cardinals.

Magidor showed that it is consistent (relative to the existence of supercompact cardinals) that the smallest strongly compact cardinal is also the smallest measurable cardinal. He also showed that it is consistent that the smallest strongly compact cardinal is the smallest supercompact cardinal. Now: Supercompact cardinals are limit of measurable cardinals, so we have a problem if we use your ordering, as the position of "strong compactness" would depend on the model we are considering, which is not what one would like, namely, that the ordering depends only on the axioms and the theories they imply.

Another example is that we may have a model with Wodin cardinals, and no strong cardinals, and models with many strong cardinals but no Woodin cardinals.

The ordering in terms of consistency strength is more subtle. It coincides (within the region where we can currently prove these things) with a natural ordering of certain inner models. Independently of whether the least strongly compact cardinal is the least measurable cardinal or not, if there is a strongly compact cardinal, there are inner models with many measurable cardinals (and much more). On the other hand, we can identify a canonical smallest inner model with a measurable cardinal (an analogue of $L$), and there can be no strongly compact cardinals in any of its submodels. This shows that in a clear sense, "strong compactness" is a stronger assumption.

The issue with Woodin cardinals and strong cardinals is similar, but there is an additional subtlety, namely, (simplifying matters a tiny bit) to be a Woodin cardinal is a "local" assumption, we just need a large enough initial segment of $V$ to verify the Woodinness of a given cardinal. However, to be a strong cardinal is a "global" assumption", that requires that we check the whole universe. Something similar occurs with supercompactness (global) and other assumptions, like the existence of rank into rank embeddings(local). What is true is that, independently of size, if there is a Woodin cardinal, then there are inner models with many strong cardinals (but, just as before, there is a canonical smallest inner model with a strong cardinal, and no cardinals are Woodin in any of its submodels), so the assumption that there is a cardinal that is Woodin is strictly stronger.

Moreover, we are more interested in "large cardinal assumptions" than in "large cardinals" per se, and for many assumptions, largeness in terms of cardinality is not a good measure. For example: The existence of $0^\sharp$ is a large cardinal assumption. It is significantly stronger than the existence of a weakly compact cardinal (for example, if $0^\sharp$ exists, then there is a proper class of weakly compact cardinals in $L$). However, $0^\sharp$ is a countable object, while weakly compact cardinals are not.

The follow up question is what large cardinal assumption is the largest (in whichever appropriate ordering we end up using). At the moment, such an assumption does not seem to exist. We have a template that generates strong large cardinal axioms, namely describing properties of elementary embeddings. We know not everything is possible: Kunen showed there are no Reinhardt cardinals, i.e., no embeddings $j:V\to V$. This gives us an upper bound on what we can have. We have been considering assumptions that in a sense look very close to this boundary (rank into rank embeddings, for example), but it may well be that closeness is a feature of the resolution of our current telescopes. Years ago, it was expected that supercompactness was an assumption just slightly above the existence of strong cardinals. Nowadays we see a huge chasm in between.