By the definition of the adjoint operator, we have $\langle Tv,w\rangle=\langle v,T^*w\rangle$. Hence $\langle T^2v,w\rangle=\langle Tv,T^*w\rangle$. Since $T$ is self-adjoint, $T^*=T$, and you get your result.
There is essentially no condition on the space on which your functions live that will ensure existence of any eigenfunctions. For example, for $X=[a,b]$ and $V$ the space of real-valued (or complex-valued, hardly matters...) functions on $X$, and $L$ the multiplication-by-$x$ operator... there are no eigenfunctions at all (unless $a=b$) [EDIT: among continuous or $L^2$ or $L^p$ functions]. That is, unless the physical space is a finite set (or has a weird-enough topology so that there aren't very many continuous functions), some very simple, non-pathological operators fail to have any eigenvalues at all (EDIT: among continuous or $L^2$ or...].
Another non-pathological example that shows that it is not generally reasonable to expect eigenvalues is the Laplacian on the real line. Fourier inversion shows that everything in $L^2$ is a superposition of generalized eigenvalues [EDIT: oop, eigenfunctions] (the exponentials), but not a sum, and those eigenvalues are not in the space itself. EDIT: Fourier inversion expresses (e.g.) a Schwartz function $f$ as
$$
f(x) \;=\; \int_{\mathbb R} e^{2\pi i\xi x}\;\widehat{f}(\xi)\; d\xi
$$
where $\widehat{f}$ is the Fourier transform of $f$ Thus, $f$ is a superposition (=integral) of eigenfunctions (the functions $x\to e^{2\pi i\xi x}$ for $\Delta=\partial^2/\partial x^2$.
Yes, in the context of Sturm-Liouville problems (see also Fredholm alternative), the point is that the inverse of the differential operator (with boundary conditions) is a compact self-adjoint operator on a Hilbert space of functions, and the eigenvalues are in bijection by $\lambda \leftrightarrow \lambda^{-1}$, etc. The only general class of operators on infinite-dimensional spaces with a clear, simple, and happy spectral theory are compact self-adjoint, or unbounded operators which have compact self-adjoint operators as inverses/resolvents...
Thus, for example, the Laplace-Beltrami operator on a compact Riemannian manifold does provably have compact resolvent, so $L^2$ has an orthonormal basis of eigenfunctions.
EDIT: I'd also add that one almost surely wants a reasonable topology on the space of functions, related to the topology on the underlying physical space. And we'd want some sort of completeness on the space of functions, else we'd potentially lose eigenfunction/values for silly reasons.
Best Answer
Long story short: there is no good way to completely "visualize" complex matrices with any useful generality. Even in the smallest non-trivial case, we are looking at a transformation over $\Bbb C^2$, which from a "geometric" standpoint is really a $4$-dimensional space.
With that said: with matrices and with other "complicated" mathematical objects, "visualization" in the usual sense is not always necessary to get a feeling for a mathematical object, and this includes complex matrices. As an analogy, I suggest you watch this video from 3Blue1Brown about 10-dimensional spheres and boxes, which are "visualized" (in a limited sense) in terms of "sliders". Note that there is really nothing geometric about a row of 10 sliders. Nevertheless, we can leverage our understanding of this representation to get a "feeling" for the fact that the volume of a box grows faster than the volume of the box's inscribed sphere as the number of dimensions is increased.
Similarly, here is a limited way in which normal matrices (which include unitary, self-adjoint, and skew-adjoint operators) can be visualized. When a real matrix is diagonalized with real eigenvalues, the picture associated with the diagonalization of a linear transformation is one of space being "stretched, squished, or flipped" along the directions corresponding to the eigenvectors of the transformation.
In the case where a complex matrix can be diagonalized with real eigenvalues (e.g. a self-adjoint operator), the span of a single vector in $\Bbb C^n$ is something that would normally be visualized as $2$-dimensional, so that the stretch/squish/flip occurs uniformly across the entirety of this "2-dimensional" complex line. With that established, we can say that the complex eigenvalue $\lambda = re^{i \theta}$ encodes an expansion by factor $r>0$ followed by a rotation by angle $\theta$ within this complex line. The spectral theorem tells us that this is enough to visualize any normal operator, and that for any normal operator, these eigenspaces will be mutually orthogonal.
With that, we can still understand the notion of independent directions and the action of the linear transformation along each of these direction, and often this is enough. What we lose, however, is our ability to visualize these directions at the same time.