Do tidal forces affect the poles? In many tidal forces illustrations it seems like the poles (with respect to the satellite's rotation) are drawn towards the center of the body. 
I can't understand why that is.
Is it because the body is assumed to be elastic to some degree? Would a small object placed at pole but not connected to it feel this increase in gravity?
 
By Krishnavedala - Own work, CC BY-SA 3.0
 A: Keep in mind these two thing:


*

*The gravitational force due to the other body points from where you measure the force toward that other body so that it has different direction depending on where you check it. 
Straight across in the vertical center of that diagram, tilting downward in the top half of the diagram and tilting upward in the bottom half.

*There arrows there don't represent the gravitational force, they represent the tidal force which is defined as the difference between the gravitational force at the point of interest and the gravitational force at the center of mass:1
$$ \vec{F}_\text{tidal}(x) = \vec{F}_\text{gravitational}(x) - \vec{F}_\text{gravitational}(x_{CoG}) \;.$$
Note that I have carefully written arrows over the forces to remind you that these are vectors and direction matter as well as magnitude.
So the tidal force at the pole at the top (bottom) of the figure is the difference of force tilted downward (upward) and a force pointing straight across; that difference points downward (upward). It also has a very small horizontal component because the distance from the pole to the other object and the distance from the CoG to the other object are almost the same so the two magnitudes are very similar.
You can follow the same kind of reasoning to explain the direction of all the arrows in the figure; just recall that the parts of the planet closer to the satellite experience a large magnitude gravitational force and those farther from experience a smaller magnitude gravitational force.

1 It might be better to deal with tidal acceleration first and then compute the tidal force as in terms of that acceleration, but that makes the convoluted language of the problem even more involved so I'm skipping it here.
A: 
In many tidal forces illustrations it seems like the poles (with respect to the satellite's rotation) are drawn towards the center of the body. I can't understand why that is.

Those places where the tidal acceleration is directed inward are not poles. They are more akin to an equator. The image in the question represents a slice. To get a picture of the tidal force over the surface of the Earth, you need to envision that picture being rotated about the line between the center of the Earth and the center of the Earth.

Is it because the body is assumed to be elastic to some degree?

There is no assumption of elasticity here. That diagram merely shows the tidal acceleration on a point on the surface of a body toward some other body ("the satellite") toward which every point on the first body accelerates due to gravitation. That second body might be "a satellite" (e.g., the Moon), but it can also be a much larger but more distance object (e.g., the Sun).
The tidal acceleration on a point on the surface of the Earth toward the Moon (or Sun) is the difference between the gravitational acceleration of a test mass at that point toward the Moon (or Sun) and the gravitational acceleration of Earth as a whole toward the Moon (or Sun):
$$a_{\text{tidal}} = GM_{\text{moon}}\left(\frac{\vec R - \vec r}{||\vec R - \vec r||^3} - \frac{\vec R}{||\vec R||^3}\right)$$
where $\vec R$ is the displacement vector from the center of the Earth to the center of the Moon and $\vec r$ is the displacement vector from the center of the Earth to the point of interest on the surface of the Earth.
This simplifies greatly in the case that $||\vec R|| \gg ||\vec r||$, which is the case for both the Moon and the Sun relative to the Earth.

Repeating the last question,

Is it because the body is assumed to be elastic to some degree?

That said, the solid Earth is elastic, and because of this, it does respond to these tidal forcing functions. Moreover, those tidal forcing functions apply all the way to the center of the Earth, but reduced in magnitude as depth increases. The response of the Earth as a whole to these inexorable forcing functions is the solid Earth tides. These Earth tides are small, about the third of the size of the ocean tides. Standing still on the surface can result in your distance from the center of the Earth varying by about 1/3 of a meter over the course of a day.
While the driving factor in the Earth tides is the vertical component of the tidal acceleration, the driving factor in the case of the ocean tides is the horizontal component of the tidal acceleration. The ocean tides are further complicated by the existence of continents, by the varying depths of the oceans, and by the responses of the oceans to those forcing functions. 
