It is helpful if you always look at the units that a certain physical quantity measures. Since you use Real-Time Rendering, I'll also quote from that (3rd edition). Also, for the sake of completeness, I'll go through all quantities and units related. I will however assume you understand solid angles. The time $t$ is measured in *seconds* $\left[s\right]$ and the solid angles $\omega$ are measured in *steradians* $\left[sr\right]$.

 1. *radiant energy $Q$* (in *joules*, $\left[J\right]$) measures the energy, i.e. the energy of a photon times the number of photons.

 2. *radiant flux $\Phi$* (in *watts*, $\left[W\right]= \left[\frac{J}{s}\right])$ measures the energy per time, e.g. don't just count the number of photons but the number of photons per second. $\Phi = \frac{Q}{t}$

 3. 
  a) *irradiance $E$* (in *watts per square meter*, $\left[\frac{W}{m^2}\right]$) measures the energy per time and surface area, or the flux per surface area. $E = \frac{Q}{t A} = \frac{\Phi}{A}$

  b) *radiosity $M$* (in some papers also $B$) is the same as irradiance, only it's leaving a surface and not arriving at it

 4. *radiance $L$* (*watts per square meter per steradian*, $\left[\frac{W}{m^2 sr}\right]$) is the radiant flux per area and solid angle, or the irradiance per solid angle. $L = \frac{\Phi}{A w}$

Now there is one thing to consider: $E$ is measured with regards to a surface $A$ that is perpendicular to the light direction (in other words, the normal of the surface is parallel to the light direction). Therefore we project $A$ onto a plane that fulfills this requirement. If the angle between the surface normal and the light direction is $\theta$, then our projected surface $A_{proj}$ is calculated thus: $$A_{proj} = \frac{A}{\cos\theta}$$

[![Surface $A$ is being projected on a plane, the result is $A_{proj}$, from "Parameterbasierte Texturgenerierung und echtzeitfähiges Rendering von nassen und trockenen Straßenoberflächen in kamerabasierten ADAS Tests", Tim Lobner][1]][1]

With this, we can of course also further our radiance calculation to $L = \frac{E}{\omega}$

But still, $E$ only considers the amount of energy per time, not from which direction it comes. Why is this important? Because of the way you usually look at lighting in computer graphics. You calculate how much light is being reflected from a surface to your viewer (/camera), which also means, that you want to know from which light source it originates (seeing as you would like to have the right amount of energy and the correct color). Additionally, you usually use point lights, meaning that you can view the lighting calculation as that of a ray from a single point (the light source) onto a single point on a surface (your pixel/fragment) and then to your viewer. These directions are written in the matter of solid angles, or to make the theory even easier, in differential solid angles. Another point is that your surface may reflect light differently depending on the where it comes from, which also makes important the direction part. 

So to sum this up a bit:

> I can't visually understand what does irradiance of a certain direction means?

It basically means photons from a specific light source, not from any place in space.

> And what is the difference with radiance of certain direction?

I hope it is clear, that radiance is irradiance from a certain direction. If not, please try to specify which part still bothers you.

> They both represents the power of light

Yes they do. In fact, *steradian* is a dimensionless unit, since it is $\left[\frac{m^2}{m^2}\right]$, and therefore it doesn't really add anything. I see how this is confusing. I hope I could clear up why you do this.

> What does irradiance of a certain direction means? Doesn't we associate a direction with radiance?

Careful. We don't associate direction (other than considering only surfaces perpendicular to the light direction) with irradiance. We do however with radiance.

> Isn't radiance defined as irradiance in a single direction?

Yes, you can say that.

> Is E(v) <= L(v) where v is direction?

I'd say it is the other way round, $L(v) \leq E(v)$, since $E(v)$ would consider any light source emitting light onto a surface, whereas $L(v)$ only considers light form $v$ (if you consider $v$ as the light source to surface direction. In my writing earlier, this is $\omega$ and in Real Time Rendering specifically, this is $\omega$ as well, or $l$ in chapter 5 as well as in the BRDF Theory chapter). Also, remember that these two physical quantities don't have the same units and should not really be compared this way.


  [1]: https://i.sstatic.net/aJrFR.png