These units look strange, because they are made up of many simpler quantities. So let's start simpler and look at velocity and acceleration first. The SI units of velocity are $[v]= m/s$, since it denotes a distance travelled over time. Acceleration is the change of velocity with respect to time so it will have units of velocity, divided by units of time $[a]= [v]/s = m/s^2$. I think this should answer your second question. If an object is accelerating at $1 m/s^2$, it means that every second, the object's velocity changes by $1 m/s$ every second. That's the meaning of $s^{-2}$.
Now, before we move on to work, let's look at Newton's second law of motion
$$F=ma.$$
Since acceleration has units of $[a]= m/s^2$, then force has units of $[F]= [m] \,[a] =kg \times m/s^2 $. This quantity is called a Newton, $N= kg\times m/s^2$. It means that an amount of force equal to $1N$ would accelerate an object of mass $1kg$ by a velocity of $1m/s$ every second.
Now we are ready to look at the definition of the Joule. Its value comes from the equation for work $W$, which is a quantity measured in joules.
$$W=Fd,$$
where $d$ is the distance travelled due to some amount of force $F$. So one can see that the units of work are $[W]= [F][d] = N\times m=kg\times m^2/s^2=J$. So you can see that the interpretation in terms of force is easy - A joule is the amount of work done by applying a force of $1N$ over a distance of $1m$. However, thinking of it in more elementary terms, things become more opaque and harder to interpret, but it's not impossible. One can also think of the joule as the energy equivalent to travelling a distance (in metres $m$) by an object of mass (in kilograms $kg$) with an acceleration $a$ (in units $m/s^2$). And this is the interpretation of $m^2$ in the definition of $J$. It has to do with an object travelling a certain distance with a certain acceleration, not an area.
Finally, the definition of power is the rate of change of work
$$P= \frac{dW}{dt}.$$
From here we can again deduce that the unit of power will have units of work (energy) divided by units of times $[P]=[W]/[t]=J/s = kg\times m^2/s^3$. One can again start to take the equation apart and think of the change in acceleration due to a force, which pushes an object to some distance over some period of time and it all becomes very complicated.
This is why we define these compound units in the first place. They save you the trouble of thinking about these ideas in an overly complicated reductionist way and they allow you to think instead about concepts such as energy, force and power, instead of objects being accelerated over a distance over a period of time...