Q1: Yes. As an example take $V=\mathbb{Z}/2\mathbb{Z} = \{ [0],[1]\}$ as a $\mathbb{Z}$-module. Then you have $$[1] \cdot 1 = 3 \cdot [1] = 5 \cdot [1] = \cdots$$
are all different representations of $[1]$ of the form $\lambda \cdot [1]$ with $\lambda \in \mathbb{Z}$.
Q2: No. (8) means that IF a module is free and is generated by $n$ elements, then every other basis has $n$ elements. (8) is not talking about modules which are not free. (3) and (6) just say that there exist some non-free modules.
The answer to your question is yes but, at least according to most treatments I know, you don't really need to know the answer to make sense of the definition of the local index. This is because the authors likely refer to the concept of "orientation-preserving" isomorphisms of oriented vector spaces from algebra rather than the "orientation-preserving" for diffeomorphisms of manifolds from geometry. The latter definition involves smoothness while the former definition doesn't. As it turns out $D_qf$ is orientation-preserving as a vector space isomorphism if and only if $D_qf$ is orientation-preserving as a diffeomorphism of manifolds, but you need an interpretation of how a vector space becomes a manifold.
To make your argument precise, the first question you need to ask yourself is how do you want to think of $T_qN$ (and $T_pM$) as a manifold? That is, what is the topology and the smooth structure on $T_qN$? Without answering this question, you can't really argue that $D_qf$ is a homeomorphism/diffeomorphism. There are at least two choices which make sense:
- Think of $T_qN$ as a vector space. Any vector space $V$ has a unique smooth structure which is obtained by declaring some isomorphism $\psi \colon \mathbb{R}^n \rightarrow V$ to be a global chart for $V$. You can check that the smooth structure doesn't depend on the choice of the isomorphism and, once you use one isomorphism, any other isomorphism will also be a global chart. If you endow two vector spaces $V,W$ with the natural smooth structures described above, you can check that any linear map $S \colon V \rightarrow W$ will automatically be smooth (in particular, continuous). Hence, if $S$ is bijective, it will be a diffeomorphism (as $S^{-1}$ is also linear, hence smooth). You can also use the fact that the differential of $S$ can be identified with $S$ itself, but it just complicates the argument. In particular, if you apply this argument to $V = T_qN, W = T_pM$ and $S = D_qf$, you'll get that $D_qf$ is a diffeomorphism.
- Think of $T_qN$ as a submanifold of the tangent bundle $TN$. One can check that $T_qN$ is indeed an embedded submanifold of $TM$ so it has a natural unique smooth structure compatible with the subspace topology, which, as it happens, turns out to be the same structure you would get if you used the vector space structure. With this interpretation, you can check that $D_qf$ is a diffeomorphism by using slice charts around $T_qN$ and $T_pM$ (which come from the construction of charts on $TN,TM$) and verifying that, in local coordinates, $D_qf$ is linear bijective map, hence a diffeomorphism. You can also argue in various other ways.
Next, in order to make sense of your interpretation, note that it is not enough to give $T_qN$ the structure of a manifold. You need to orient it as well. How you will do that depends on your definition of orientation (as there are many equivalent definitions). If an orientation is defined by giving an oriented atlas, the easiest thing to do is to work with the first interpretation above. If $X \colon U \rightarrow N$ is an oriented chart around $q$ with $X(a) = q$, define an oriented smooth structure on $T_qN$ by declaring the differential $DX|_a \colon T_a(\mathbb{R}^n) \rightarrow T_qN$ to be an oriented chart (where you identify $T_a(\mathbb{R}^n)$ with $\mathbb{R}^n$ in the usual way). If your definition of orientation is different, you might need to do something different.
As you can see, there are many details to fill in order to work with your interpretation. However, most books I know (I haven't checked Tu nor Marsden) also discuss the notion of an orientation of a vector space which is a pure linear algebra notion unrelated to any issues of smoothness. Then one defines when a map between oriented vector spaces is orientation preserving and finally, one shows that the definition of orientation on a manifold $N$ induces an orientation for each tangent space $T_qN$ (which "varies smoothly" with respect to $q$). Then, the definition of index is with respect to the notion of orientation preserving/reversing linear maps between oriented vectors spaces and not diffeomorphisms between oriented manifolds. This gives a conceptually cleaner treatment as it seperates the issue of smoothness from the issue of being orientation preserving/reversing.
Best Answer
For any integer $k$, the set $M_k$ of complex-differentiable functions $f$ defined on the upper-half plane $\{x+iy: \, y > 0\}$ that satisfy the equations $$f(z+1) = f(z), \; \; f(-1/z) = z^k f(z)$$ and have limit $\lim_{y \rightarrow \infty} f(iy) = 0$ is a vector space over $\mathbb{C}$.
Two specific elements of $M_k$ include the functions $$E_4(z) = 1 + 240 \sum_{n=1}^{\infty} \sigma_3(n) e^{2\pi i n z} \in M_4$$ and $$E_8(z) = 1 + 480 \sum_{n=1}^{\infty} \sigma_7(n) e^{2\pi i nz} \in M_8.$$ Here, $\sigma_k(n)$ is the divisor sum $\sum_{d | n} d^k$.
Assuming that $E_4 \in M_4$ it is rather easy to show that $E_4^2 \in M_8.$
It can be proved that $M_8$ is one-dimensional, so $E_4^2$ is a multiple of $E_8$. Comparing constant coefficients tells you that they must be equal, and comparing the others gives you the formula $\sigma_7(n) = \sigma_3(n) + 120 \sum_{m=1}^{n-1} \sigma_3(m) \sigma_3(n-m).$
For example $$\sigma_7(2) = 1 + 2^7 = 1 + 2^3 + 120$$ and $$\sigma_7(3) = 1 + 3^7 = 1 + 3^3 + 120(1+2^3 + 1 + 2^3).$$
A lot of vector spaces like this show up in number theory. They are typically finite-dimensional but working out a basis is pretty hard (certainly harder than showing that they are finite-dimensional).