Spectral decomposition of $-\Delta$ the Laplacian
A confession: I can only really answer this when the manifold in question is an open set $U$ of $\mathbb{R}^n$, where $\partial U$ is smooth. (I would imagine this case is used, or at least very instructive for, the case on a general Riemannian manifold.) But I'll try to answer in this setting as best I can. I apologize in advance for the length of this answer.
2) The first thing that must be done is to understand what $V$ is, because functional analytic machinery needs to be applied on a particular (Hilbert) space. Indeed, in this case $V = H^1_0(U)$. I don't know if there is any way to get around having to deal with this object, but you can think of $H^1_0(U)$ as the space of all function $f$ which are zero on the boundary (this is what the "$0$" in the subscript refers to), are square-integrable, and whose derivatives exist and are also square-integrable. This is a Hilbert space. The inner product is not just the $L^2$ inner product anymore, because the inner product on $L^2$ does not care about the derivatives, but our space does. So we define the inner product by $$ \langle f, g \rangle = \int_U f(x)g(x)\, dx + \int_U \nabla f(x) \cdot \nabla g(x)\, dx. $$ Note that $\langle f, g \rangle = \langle f, g\rangle_{L^2} + \langle \nabla f, \nabla g\rangle_{L^2}$. The last thing that must be done is to note that this space is not complete if we restrict to the usual defintion of derivatives, and therefore the formal definition is that $f, g$ have square-integrable "weak" derivatives. This is morally perhaps unimportant, but important in the rigorous proofs. This is essentially why you cannot take $V = C^2(U) \cap C^2(\bar{U})$, since this space is not complete under the norm induced by the inner product above, and so if we take a limit of functions in this space, the result won't necessarily be $C^2$.
1) Your calculations are essentially correct, but can be (and in the general theory are) extended to functions with these "weak" derivatives. It turns out not to be important for your calculations involving $\phi_i$ which are eigenvalues of $-\Delta$, since elliptic regularity actually implies all eigenfunctions of $-\Delta$ are smooth.
3) This is where the full machinery comes in. As rubikscube09 mentioned in the comments, I don't believe there is a way to do this without the Rellich-Kondravich theorem (all of this is in Chapter 5, 6 of Evans) and abstract functional analysis. Essentially, the R-K theorem says that the embedding $H^1_0(U) \to L^2(U)$ is compact, in that bounded sequences in $H^1_0(U)$ have an $L^2$-convergence subsequence. Therefore we may think of $(-\Delta)^{-1}$ as mapping from $L^2$ to $H^1_0$ defined by $(-\Delta)^{-1}f = u$ is the unique function such that $-\Delta u = f$. Then $u$ lives in $H^1_0(U)$ (and again, there is something being swept under the rug, since $u \in H^1_0(U)$ only implies $u$ has first order derivatives, not second-order), but the embedding $H^1_0(U) \to L^2(U)$ identifies $u$ as an $L^2$ function in a compact way and thus we may think of $(-\Delta^{-1}) : L^2 \to L^2$. By R-K this is a compact operator.
4) You addressed this in your comment.
5) This is a computation that is a little involved, but not too long. I don't know of free-source materials, but Evans ch. 6 deals with this (in more generality) and has detailed proofs there. There may be copies online?