Have you studied Linear Algebra? That's really where it comes from. If A is a linear operator on some vector space, the "eigenvalue problem" is: [itex]Av= \lambda v[/itex]. That has, of course, the "trivial" solution v= 0. For some values of [itex]\lambda[/itex], called "eigenvalues", there are other, non-trivial, solutions- in fact the set of all solutions, in that case, is a sub-space. Those non-trivial solutions are the "eigenvectors".
Of course, in the finite dimensional case, after we have chosen a basis, we can write a linear operator as a matrix, with each column showing what the operator takes each basis vector to (written in terms of that basis). If we choose an eigenvector, corresponding to eigen value [itex]\lambda[/itex] as the "nth" basis vector, the nth column consists of "0"s except for the value [itex]\lambda[/itex] in the nth row. In particular, if we can find a "complete set of eigenvectors"- that is, a basis consisting entirely of eigenvectors- which we always can in the important "self adjoint operator" case, then the matrix representing the linear operator in that basis is diagonal- the simplest kind of matrix.
If the vector space is a "function space", typically, infinite dimensional, we cannot write the operator as a matrix but still, using as many eigenvectors (eigenfunctions in this case) as basis vectors, and especially in the "self adjoint" case, using eigenfunctions for all basis functions, allows us to write the operator in a particularly simple form.
As for the "physical significance"- like any question about the physical significance of a mathematical concept, that depends entirely upon the particular physics application and which "model" you are using. That would be a question more appropriate to a physics forum that a mathematics forum.