This is a big topic with many aspects but let me start with the reason why entropy and the second law was needed.
You know the first law is conservation of energy. If a hot body is placed in contact with a cold body heat normally flows from the hot body to the cold . Energy lost by the hot body equals energy gained by the cold body. Energy is conserved and the first law obeyed.
But that law would also be satisfied if the same amount of heat flowed in the other direction. However one never sees that happen naturally (without doing work). What's more, after transferring heat from hot to cold you would not expect it to spontaneously reverse itself. The process is irreversible.
The Clausius form of the second law states that heat flows spontaneously from hot to cold. Clausius developed the property of entropy to create this as a general state function that could eventually be determined independently of trying to map just heat flow.
ADDENDUM 1:
Found a little more time to bring this to the next level. This will tie in what I said above to the actual second law and the property of entropy.
So we needed a new law and property that would be violated if heat flowed naturally from a cold body to a hot body. The property is called entropy, $S$, which obeys the following inequality:
$$\Delta S_{tot}=\Delta S_{sys}+\Delta S_{surr}≥0$$
Where $\Delta S_{tot}$ is the total entropy change of the system plus the surrounding (entropy change of the universe) for any process where the system and surroundings interact. The equality applies if the process is reversible, and the inequality if it is irreversible. Since all real processes are irreversible (explained below), the law tells us that the total entropy of the universe increases as a result of a real process.
The property of entropy is defined as
$$dS=\frac {dQ_{rev}}{T}$$
where $dQ$ is a reversible differential transfer of heat and $T$ is the temperature at which it is transferred. Although it is defined for a reversible transfer of heat, it applies to any process between two states. If the process occurs at constant temperature, we can say
$$\Delta S=\frac{Q}{T}$$
where $Q$ is the heat transferred to the system at constant temperature.
We apply this new law to our hot and cold bodies and call them bodies $A$ and $B$. To make things simple, we stipulate that the bodies are massive enough (or the amount of heat $Q$ transferred small enough) that their temperatures stay constant during the heat transfer Applying the second law to our bodies:
$$\Delta S_{tot}=\frac{-Q}{T_A}+\frac{+Q}{T_B}$$
The minus sign for body $A$ simply means the entropy decrease for that body because heat is transferred out, and the positive sign for body $B$ means its entropy has increased because heat is transferred in.
From the equation, we observe that for all $T_{A}>T_{B}$, $\Delta S_{tot}>0$. We further note that as the two temperatures get closer and closer to each other, $\Delta S_{tot}$ goes to $0$. But if $T_{A}<T_{B}$ meaning heat transfers from the cold body to the hot body, $\Delta S$ would be less than zero, violating the second law. Thus the second law precludes that natural transfer of heat from a cold body to a hot body.
Note that for $\Delta S_{tot}=0$ the temperatures would have to be equal. But we know that heat will not flow unless there is a temperature difference. So we see that for all real heat transfer processes, such processes are irreversible.
Irreversibility and entropy increase is not limited to heat transfer processes. Any process goes from a state of disequilibrium to equilibrium. Beside heat, you have processes involving pressure differentials (pressure disequilibrium). These process are also irreversible and generate entropy.
ADDENDUM 2:
This will focus on the specific questions no. 1 and 2 in you post, that is
1. A process has an entropy of X what does this tell me?
2. Another process has higher entropy what does this tell me?
Before answering this, it has been said that when the change in entropy, $\Delta S$, is positive, “heat has entered the system”. It should be noted that heat entering the system is a sufficient condition for a positive entropy change, but it is not a necessary condition.
As I said above, irreversibility and entropy generation is not limited to heat transfer processes. For example, an irreversible adiabatic expansion results in an increase in entropy, although no heat transfer occurs.
An example is the free adiabatic expansion of an ideal gas, a.k.a. a Joule expansion. A rigid insulated chamber is partitioned into two equal volumes. On one side of the partition is an ideal gas. On the other side a vacuum. An opening is then created in the partition allowing the gas to freely expand into the evacuated half. The process is irreversible since the gas will not all return to its original half of the chamber without doing external work (compressing it).
Since there was no heat transfer between the gas and the surroundings, $Q=0$, and since the gas expanded into a vacuum without the chamber walls expanding, the gas does no work, $W=0$. From the first law, $\Delta U=Q-W=0$. For an ideal gas, any process, $\Delta U=C_{v}\Delta T$. Therefore there is no change in temperature. The end result is the volume of the gas doubles, the pressure halves, and the temperature remains the same.
We can determine the change in entropy for this process by devising a convenient reversible path to return the system to its original state, so that the overall change in entropy for the system is zero. The obvious choice is a reversible isothermal (constant temperature) compression process. The work done on the case in the isothermal compression equals the heat transferred out of the gas to the surroundings (increasing its entropy) and the change in internal energy is zero. Since this occurs at constant temperature we have, for the gas (system),
$$\Delta S=-\frac{Q}{T}$$
Since we have returned the system to its original state, the overall change in entropy of the system is zero. Therefore, the change in entropy due to the free expansion had to be
$$\Delta S_{exp}=+\frac{Q}{T}$$
We could also determine $\Delta S$ by combining the first law and the definition of entropy. This gives the second equation in Jeffery’s answer, which for the case of no temperature change ($dT=0$) gives us, for one mole of an ideal gas,
$$\Delta S=R\ln\frac{V_{f}}{V_i}$$
or, in the case of our free expansion where the volume doubles,
$$\Delta S=R\ln2$$
Therefore,
$$\Delta S=\frac{Q}{T}=R\ln2$$
Now, to answer your questions, what does this tell us? And what does another process having higher entropy tell us?
Or, to put it another way, why should we care?
One thing it tells us is that, in the case of an ideal gas, an irreversible (free) adiabatic expansion of an ideal gas results in a lost opportunity to do work. In the free adiabatic expansion, no work was done. If, however, the process was a reversible adiabatic process against a variable external pressure (constant entropy process), such that $Pv^k$=constant ($k=\frac{C_{p}}{C_{v}})$ the gas would have performed work on the surroundings equal to
$$W=\frac{(P_{f}V_{f}-P_{i}V_{i})}{(1-k)}$$
Bottom line: One of the ramifications of an irreversible expansion process is that the work performed will be less than that for the same process carried out reversibly, due to the generation of entropy in the irreversible process. Irreversible processes lower the thermal efficiency of a system in performing work.
Hope this helps.
Best Answer
Arnold Neumaier's answer is correct but doesn't seem to have included enough detail to convince people, so here is an answer with a more in-depth explanation.
We have two fundamental theories of physics: general relativity and the standard model of particle physics. The standard model has CPT symmetry, and general relativity has local time-reversal invariance. Although neither of these is technically the same as global time-reversal invariance, for the purposes of the following discussion it's sufficiently accurate to say that the laws of physics are time-reversal invariant. Sometimes you will hear people state this by saying that the "microscopic laws" are time-reversal invariant, the intention presumably being to exclude the second law of thermodynamics, which explicitly distinguishes a forward direction in time. But this is an anachronism, since the second law is no longer considered fundamental but derived.
The question that then arises is, how in the world can you derive a time-asymmetric theorem from time-symmetric assumptions?
Consider the simulation shown below. On the right we have a box that has three areas marked with three colors, and $N=100$ particles that are free to move around in the whole box. (The vertical lines at the boundaries are just visual -- the particles cross them freely.) The simulation was done using this applet. The particles are released at random positions, with random velocity vectors, and their motion is simulated using Newton's laws, which are time-reversal symmetric. The graph on the left shows the number of particles in each area as a function of time.
Since the particles are initially placed randomly, roughly one third of them are initially in each region. At any randomly chosen time, the number of particles $n$ in, say, the red region has a mean of $\bar{n}=N/3$ and a standard deviation of about $\sqrt{\bar{n}}\approx 6$. Once in a while we get unusually large fluctuations, such as the one marked with a green arrow at $t=19$.
We can now state a derived law L:
(L) If we observe a statistically unlikely value of $n$ at some time $t_0$, there is a high probability that the values of $n$ both before and after $t_0$ (for $t\lesssim t_0-3$ and $t\gtrsim t_0+3$) are closer to the mean.
As $N$ gets larger and larger, L becomes more and more secure; the probability of seeing it violated becomes smaller and smaller. When $N$ gets as big as Avogadro's number, the probability of a violation becomes zero for all practical purposes.
This derived law is still completely time-reversal symmetric, so it doesn't appear to be quite the same as the second law of thermodynamics. But now consider the case where somebody artificially prepares the particles in the box so that they are all initially in the center. (If you run the applet at the link above, this is actually what it does.) The result is shown below.
An observer who doesn't know about the initial preparation of the system, and who only gets to see its behavior during the interval $0\lt t \lesssim 2$, will empirically arrive at a time-asymmetric "law" describing the behavior of the system: the system always evolves from high values of $n_{\text{black}}$ to lower ones. Not knowing the initial preparation of the system, but wishing to believe in a naturalistic theory of the operation of this little "universe," the observer might speculate that the initial, high value of $n$ was an extreme statistical fluctuation. Perhaps at $t\lesssim -2$ the system was in equilibrium. The observer can then explain everything in terms of the time-symmetric law L.
The same analysis applies to the conditions we observe in our universe, with some modifications:
The discussion in terms of $n$ can be replaced with a discussion in terms of the number $\Omega$ of accessible states for a given set of coarse-grained observables -- or we can talk about $\ln\Omega$ or $k\ln\Omega$, i.e., entropy, which is additive.
In the original statement of L we had a constant time 3, which was an estimate of the equilibration time for the toy model. For the real universe, this has to be replaced by some estimate of the equilibration time of the whole universe, which might be very long.
And finally, we have the role played by the mischevious person who secretly initialized the system with all the particles in the center. This naughty trickster was effectively setting a boundary condition. In our universe, this boundary condition consists of the fact that, for reasons unknown to us, our Big Bang had a surprisingly low entropy. If there were some naturalistic principle that the Big Bang should be a typical state rather than a very special one, then our universe should have started out already in a state of maximum entropy.
In the world around us, we see various arrows of time. There is a psychological arrow (we can remember the past, but not the future), a thermodynamic one (candles burn but never unburn), a cosmological one (the Big Bang was in our past, not our future), and various others such as a radiative one (we often observe outgoing spherical radiation patterns but never their time-reversed versions). All of these arrows of time reduce to the cosmological one, which arises from a boundary condition.
The OP asked:
No. In fact, the world that is overwhelmingly the most probable and natural is one in which the entropy is, always has been, and always will be the maximum possible -- but in such a universe there would not be hairless primates tapping on computer keyboards. It is also certainly possible to have a universe in which entropy is always higher in the past and lower in the future. In fact, our own universe is an example, if we simply interchange the arbitrary labels "past" and "future."
A longer discussion of these ideas, with lots of historical context, is given in Callender 2011. Historically, there has been a lot of debate and confusion on these issues, and unfortunately you will hear a lot of this confusion echoing down the halls a hundred years later, perhaps due to the tendency of textbooks to hew to tradition. For example, Ritz and Einstein had a debate in 1909 on the radiative arrow (as discussed in Callender and references therein). Ritz's position, that the radiative arrow is fundamental, is no longer viable.
References
Callender, Craig, "Thermodynamic Asymmetry in Time", The Stanford Encyclopedia of Philosophy (Fall 2011 Edition), Edward N. Zalta (ed.), http://plato.stanford.edu/archives/fall2011/entries/time-thermo