Is there any correlation between eigen functions and eigen vectors of an LTI system?
Then why do we say a second order LTI system has only 2 eigen
  values/vectors?"


Consider the following 2nd order system:

$$a\ddot y(t) + b\dot y(t) + cy(t) = x(t)$$

We can write this as

$$\mathbf Dy(t) = x(t) $$

where \$\mathbf D\$ is the 2nd order operator

$$\mathbf D = a\frac{d^2}{dt^2} + b\frac{d}{dt} + c$$

We know that \$e^{st}\$ is an eigenfunction of the operator \$\mathbf D\$ with eigenvalue \$\lambda = as^2 + bs + c\$

$$\mathbf De^{st} = \lambda e^{st} = (as^2 + bs + c)e^{st}$$

thus there are an infinity of independent eigenfunctions for this linear time invariant differential operator.

Next, consider the equation

$$\mathbf Dy(t) = 0$$

Now we have a constraint - we seek the specific eigenfunction(s) with eigenvalue \$\lambda =0\$.

As is well known, we proceed as follows:

$$\lambda = as^2 + bs + c = 0$$

which has two independent solutions

$$s = s_{0\pm} = -\frac{b}{2a} \pm \sqrt{\frac{b^2}{(2a)^2} - \frac{c}{a}}$$

These particular eigenfunctions are clearly special in that they are the modes of the system - the form of the system output when there is no input.