Entropy is not a force, no. It is a "chaos factor", if you will. The more entropy, the less structured a system is.
A system will always move towards the state that is most probable. The most probable state (macro-state) will be the state with most micro-state configurations.
Consider as an example four coins of heads H and tails T. Flip them and you can have the following five outcomes:
1: HHHH
2: HTHH, HHHT, HHTH, THHH
3: HHTT, HTHT, HTTH, THTH, TTHH, THHT
4: HTTT, THTT, TTTH, TTHT
5: TTTT
The five different results corresponds to five different macro-states. And they are all equally likely.
Macro-state 3 is most probable since it happens for most micro-state configurations. So, if your outcome is macro-state 3, and you flipped the coins without keeping track of each of them, then you now know the least about the micro-state configuration because there are six different possibilities of which coins gave which result. If the outcome is macro-state 1 then you know very much about the micro-state configuration because there is way fewer possibilities.
In macro-state 3 we have more chaos or higher entropy. If you make the coins flip 100 times, then it is highly likely that you reach macro-state 3 in most of the cases. It tells us that the system tends to end at the macro-state where the entropy or the chaos is higher, because that is the same as saying the macro-state with most possible micro-states and therefor the macro-state that will enclose most of the outcomes.
I've seen the example of an ice cube or a cup of hot tea, but I don't quite see how these two relate to entropy
If you have an ice cube then the structure is more "ordered" the colder it gets and less ordered if it melts. Simply because you know less and less about the position of atoms and particles within this structure the hotter it gets or the more loose they are bound.
If an ice cube melts, then it has gained entropy. The room around it has lost entropy. The entropy for either ice cube or room can be calculated from the heat energy added and the temperature:
$$S=\int \frac{1}{T}\mathrm dQ.$$
Total entropy $\sum S=S_{object}+S_{surroundings}$ turns out to be never smaller than zero. This is the 2nd law of thermodynamics. The total entropy will always increase (or be zero for a perfectly reversible process) in any process that happens.
The cup of coffe is the other way around. The coffe looses heat, and the surroundings gain that same heat. Entropy is lowered in the coffee and increased for the surroundings, but will overall in total be positive.