(1) The virtual dimension is only equal to $-i(\xi)$ once you have chosen enough trivializations and so forth that it works out that way. This formula hides that this choice of trivialization is what sees $\Lambda$. Take a look at, for instance, the Bourgeois-Ekholm-Eliashberg paper on the surgery formula for contact homology for a discussion of choice of trivializations. I will give a very fast summary. (It is also discussed in Eliashberg-Givental-Hofer's introduction to SFT, and in a number of other papers on the subject of contact homology etc.)

The Conley-Zehnder index is associated to a path of symplectic matrices starting at the identity and ending at a matrix without any eigenvalue equal to $1$. In order to obtain a path of matrices from a periodic orbit, you must trivialize $TM$ over $\xi$. In particular, the Conley-Zehnder index depends on trivialization.

The general index formula, however, has a Conley-Zehnder index term *and* a Maslov index term (and a relative Chern number term, if your boundary trivializations don't extend to a trivialization of $u^*TM$). The CZ index and the Maslov index each depend on trivialization. The difference/sum does not.

In this case, what you have implicitly done is trivialized $u^*TM$ so that the Lagrangian loop $u|_{\{0\}\times S^1}^*TL$ in $u|_{\{0\}\times S^1}^*TM$ has Maslov index $0$. Note that $u^*TM$ is trivial because $u$ is a surface with boundary. Your choice of trivialization on the boundary now extends to one of $u^*TM$. Finally, by the asymptotic convergence theorem, this induces a trivialization of $\xi^*TM$. You use this trivialization to compute the Conley-Zehnder index.

(2) Orbicular is correct. You can see the problem in a very simple toy problem. Consider finding holomorphic maps from the disk $D$ to $\mathbb{C}$. A totally real (actually Lagrangian) boundary condition is given by asking the boundary of the disk to map to the unit circle -- this has a 3 dimensional solution space. Let's take $P = \mathbb{C}$, say. Start with the standard map on the disk given by $u(z) = z$. Now take any simple closed curve in $\mathbb{C}$ that is close to the unit circle. Then, there exists a 3-parameter family of holomorphic disks with boundary on that curve. The space of such curves is infinite dimensional, showing that my linearized problem has an infinite dimensional kernel.

To see the problem for $P$ of lower dimension, we want to look in $\mathbb{C}^2$. Here, choose a generic circle for the boundary condition. For general choice of circle, you get a boundary value problem you cannot solve. You can work out by hand the infinite codimension condition you get on the circle that actually admits any solutions at all.