I feel a little lost. I think my problem is quite simple, but unfortunately, my knowledge of math is also simple.
I have two rigid bodies that move relative to each other. I know two points of each body (see picture). 
How can I find the place or point where the lower body rotates against the upper body? My idea was to just connect the points of each body and the intersection is the point I am looking for. 
but this doesnt seem right.
Thank you in advance for your help, I really apreciate it.
What you might be looking for is the relative instantaneous centre of rotation (ICR for short) of the two bodies. This is the point in space that, at a given point in time, one body will rotate about the other body, as though there is a pivot connecting the rigid bodies. The reason why this is called "instantaneous" is because, for a general arbitrary motion of two rigid bodies, this apparent pivot may change its location on either body as time progresses (note: the relative centre of rotation need not be located physically on either body).
(this answer only considers the 2D case, i.e. planar motion. This answer doesn't apply to the more general case of spatial motion, where not all velocity vectors lie within a single plane)
An illustrative example that shows how an instantaneous centre of rotation might change location on either body is the rolling of a ruler over the curved part of a fixed semi-circle without slip:
Three different points in time are shown, and it can be seen that the instantaneous pivot , i.e. the ICR, occurs at the point of contact. However, at different times, the pivot has changed location, with the coloured segments showing how much the ICR has moved on each body since the first point of time.
However, if there is a physical pivot directly linking the two rigid bodies, then the relative centre of rotation does not change location on either body, and so the term "instantaneous" is not necessary. The presence of a such a pivot restricts the possible relative motions of the rigid bodies!
So, how is the relative ICR defined mathematically? In the general case of two rigid bodies moving, there is one and only one point at which the velocity of body 1 and the velocity of body 2 are the same: this point is the relative ICR. (If body 2 is not rotating with respect to body 1, then the ICR cannot be evaluated: how can there be a centre of rotation without any rotation, after all?).
To find the relative ICR of the rigid body pair, we need to find the absolute instantaneous centres of rotation for the individual bodies. The (absolute) ICR of body 1 is defined as the point on body 1 where there is no velocity. Similarly, the (absolute) ICR of body 2 is defined as the point on body 2 where there is no velocity. The absolute ICRs are easy to find, since it is the point to which all velocity vectors lie perpendicular to. Therefore, to find the (absolute) ICR for a body, we need to know the velocity vectors at two points on that body. Then, for each of those two points, the (absolute) ICR will lie on the line passing through the point perpendicular to the direction of velocity. We can therefore draw two lines and the (absolute) ICR must lie on the intersection of those lines. If not, the ICR would not lie perpendicular to one of the velocities.
The following diagram shows the case in the question, with velocities that I random assigned to each of the four points $P_1$, $P_2$, $P_3$, $P_4$, such that the magnitudes of their velocities are $V_1$, $V_2$, $V_3$, $V_4$ respectively.
In the diagram, lines are drawn perpendicular to the velocity vectors, and the intersections of the lines reveal the absolute ICRs, labelled as points $I_1$, $I_2$ for bodies 1 and 2 respectively.
With the positions of the absolute ICRs determined, we can now measure the distances between the points at which velocities are known, and the absolute ICRs: these distances are represented in the diagram by $a_1$, $a_2$, $a_3$, $a_4$.
We know from circular motion that the tangential velocity is equal to the angular velocity multiplied by the distance from the pivot, and so the absolute value of the angular velocities of each body can be determined with
$$|\omega_1| = \frac{V_1}{a_1} = \frac{V_2}{a_2}$$
$$|\omega_2| = \frac{V_3}{a_3} = \frac{V_4}{a_4}$$
Note that, for each body, both fractions must give the same angular velocity magnitude. Otherwise, the velocities of the body are not valid for rigid body motion. (Consider the case of two points on the same body moving in opposite directions: this would suggest the body is deforming, which contradicts the assumption of a rigid body)
The sign of the angular velocities can be established by using a sign convention. Here, I'll use the convention where anticlockwise angular velocities are postive, so that
$$\omega_1 = |\omega_1|$$
$$\omega_2 = -|\omega_2|$$
Now, to determine our relative ICR, the point at which the velocities of both bodies are equal, it is important to note that the relative ICR must lie on the line connecting $I_1$ and $I_2$. This is because the velocities of both bodies considered at the same point can only be parallel to one another along this line, due to the velocities being perpendicular to $I_1$ and $I_2$. The following diagram shows the velocities at a point on the line and another point off the line to illustrate this.
So, the relative ICR must lie on this line, but where on this line? Note that as the point gets further from $I_1$, the velocity of body 1 at that point increases proportionally. Similarly, as the point moves away from $I_2$, the velocity of body 2 at that increases proportionally. Therefore, there must be a point at which the two velocities become equal.
Let $I_{21}$ represent the point of relative ICR. We can determine the length of line connecting $I_1$ and $I_2$, which we will represent as $L$. Then, let's define $l_1$ as the distance between $I_1$ and $I_{21}$, and $l_2$ the distance between $I_2$ and $I_{21}$. Note that $l_1 + l_2 = L$. Calculating the value for $l_1$ and $l_2$ will then determine the position of $I_{21}$
To determine $l_1$ and $l_2$, we use the fact the velocity $V_{21}$ at $I_{21}$ must be the same:
$$V_{21} = l_1 \omega_1 = -l_2 \omega_2$$
(This expression holds provided a sign convention is used where both angular velocities have the same direction if they have the same sign.)
From this, it can be seen that
$$l_1 = \frac{\omega_2}{\omega_2 - \omega_1}L$$
$$l_2 = \frac{\omega_1}{\omega_1 - \omega_2}L$$
(For those familiar with the radii of the pitch circles of gears, these expressions might be familiar!)
(It can be seen that both $l_1$ and $l_2$ lie in the range $0 < l_{1,2} < L$ if and only if $\omega_1$ and $\omega_2$ has different signs.)
Now the position of $I_{21}$ is revealed. Note however that the relative ICR has been calculated only for a single instance in time. To calculate for other points in time, this above must be repeated using the velocities of points $P_1$, $P_2$, $P_3$, $P_4$ at that time.
In vector notation
Note that the above can be adapted into vector notation. This way, it is useful to consider a point $P_1$ on body 1 with position $\mathbf{R}_1$ and known velocity $\mathbf{V}_1$, and body 1 has known angular velocity $\omega_1\mathbf{k}$.
$P_2$, $\mathbf{R}_2$, $\mathbf{V}_2$, $\omega_2\mathbf{k}$ are defined similarly for body 2.
The velocity $\mathbf{v}_1$ of body 1 at an arbitrary point $\mathbf{r}$ can be expressed as
$$\mathbf{v}_1 = \mathbf{V}_1 + \omega_1 \mathbf{k} \times \left(\mathbf{r} - \mathbf{R_1} \right)$$
$\mathbf{v}_2$ of body 2 at position $\mathbf{r}$ is similarly expressed as
$$\mathbf{v}_2 = \mathbf{V}_2 + \omega_2 \mathbf{k} \times \left(\mathbf{r} - \mathbf{R_2} \right)$$
By cross-multiplying these expressions with unit vector $\mathbf{k}$, the positive vectors of the absolute ICRs are seen to be equal to
$$\mathbf{I}_1 = \mathbf{R}_1 + \frac{1}{\omega_1} (\mathbf{k} \times \mathbf{V}_1)$$
$$\mathbf{I}_2 = \mathbf{R}_2 + \frac{1}{\omega_2} (\mathbf{k} \times \mathbf{V}_2)$$
The position of the relative ICR is given by
$$\mathbf{I}_{21} = \frac{\mathbf{k} \times (\mathbf{V}_2 - \mathbf{V}_1) + \omega_2 \mathbf{R}_2 - \omega_1 \mathbf{R}_1}{\omega_2 - \omega_1}$$
It depends on how those two rigid bodies are moving. Are they turning around a common center, or are they orbiting around in an ellipse? if it is an ellipse, is this ellipse wobbling, or may be they are being attracted by the force of a field which changes by coordinate and time. What is the cause of this rotation, magnetic field, or they are floating in a whirlpool, or gravity or tied by a wire?
Assuming they are just simply rotating around a common center are the facing each other continually or they have their different rotation period.
in the most basic case they could be rotating around a common point but like Earth and Mars they have different orbits.
You need some basic information before you can start to draw lines.
