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.
Let's just work in $\mathbb{R}^n$ to try to get some geometric intuition. I will directly connect the intuitive definition with the abstract one.
First, let's actually start with $n=3$. In $\mathbb{R}^3$, we can visually a vector at a point $p$ as an arrow starting at $p$, where the direction of the arrow is based on the coordinates of the vector. Visually, we view the tangent plane to $p$ as a plane the touches a surface only at the point $p$. Then, tangent vectors to this surface are all of the "arrows" starting at $p$ that lie in this plane. Note that this requires us viewing our space as being embedded in $\mathbb{R}^3$.
This concept doesn't carry over to manifolds very well. Instead, we will construct an equivalent characterization which does. The tangent space $T_p\mathbb{R}^n$ to $\mathbb{R}^n$ at a point $p\in\mathbb{R}^n$ consists of all arrows starting at $p$. If we view tangent vectors this way then we get an isomorphism between the tangent space and $\mathbb{R}^n$ by sending arrows to column vectors. For now, we'll endow the tangent space with the standard basis $e_1,\cdots, e_n$. Let $p=(p^1,\cdots, p^n)$ be a point and $v=\langle v^1,\cdots, v^n\rangle $ be a vector, all in $\mathbb{R}^n$ (the bracket notation to distinguish a vector from a point). The line through $p$ with direction $v$ is $c(t)=(p^1 +tv^1,\cdots, p^n+tv^n).$ If $f$ is smooth on a neighborhood of $p$, then we can define the directional derivative of $f$ in the direction of $v$ at $p$ to be $$D_v f=\frac{d}{dt}\Big|_{t=0}f(c(t)).$$ Via the chain rule, $$D_v f=v^i\frac{\partial f}{\partial x^i}(p),$$ which is a number, not a function. We can write $$D_v=v^i \frac{\partial }{\partial x^i}\Big|_p,$$ which takes a function to a number. The map $v\mapsto D_v$, which sends a tangent vector to an operator on functions, will give a useful way to describe tangent vectors.
First, we define an equivalence class of smooth functions in a neighborhood of $p$ as follows: consider pairs $(f,U),$ where $U$ is a neighborhood of $p$ and $f\in C^\infty(U).$ We define the relation $\sim$ as $(f,U)\sim (g,V) $ if there exists an open set $W\subset U\cap V$ containing $p$ so that $f=g$ on $W$. An equivalence class of $(f,U)$ is called a germ of $f$ at $p$, and we write $C_p^\infty(\mathbb{R}^n)$ to be the set of all germs of smooth functions on $\mathbb{R}^n$ at $p$, which forms an $\mathbb{R}$-algebra.
Now, given a tangent vector $v$ at $p$, we have $D_v:C_p^\infty\rightarrow\mathbb{R}$. It's not hard to obtain that $D_v$ is linear and satisfies the Leibniz rule: $$D_v(fg)=(D_vf)g(p)+f(p)D_vg.$$ A map $C^\infty_p\rightarrow\mathbb{R}$ that has these two properties is called a point derivation of $C_p^\infty.$ Call this set $\mathcal{D}_p(\mathbb{R}^n)$, which forms a vector space.
Okay, so we've defined a bunch of stuff, but it turns out to be very useful. We have that all directional derivatives at $p$ are derivations at $p$. So, we have a map $\Phi:T_p\mathbb{R^n}\rightarrow \mathcal{D}_p(\mathbb{R}^n)$ given by $v\mapsto D_v,$ as hinted at earlier. By linearity of $D_v$, this is a linear map. It does not take too much work (I can add, if necessary), that this map is actually an isomorphism between $T_p\mathbb{R}^n$ and $\mathcal{D}_p(\mathbb{R}^n)$. Under this isomorphism, we can identify the standard basis $e_1,\cdots, e_n$ for $T_p\mathbb{R}^n$ with the set $\partial/\partial x^1\big|_p,\cdots,\partial/\partial x^n\big|_p.$ That is, if $v$ is a tangent vector, then we can write $v=v^ie_i$ as $$v=v^i\frac{\partial}{\partial x^i}\Big|_p.$$ So, we can define tangent vectors in a geometric way in $\mathbb{R}^n$, and it turns out to be equivalent to the derivation definition, where as you said, tangent vectors are operators (specifically linear functionals on $C_p^\infty$). This definition extends way more naturally to general manifolds than the "arrow" one.
This follows the construction from Tu's "Introduction to Manifolds" fairly closely, which you might like to check out, as a further reference.
Best Answer
It might be easier to see if we put this in the context of single variable calculus.
In calculus, given a functions $f:\mathbb R\to\mathbb R$, we can take the derivative at a single point $p\in\mathbb R$. This gives us the line $$f'(p)=\frac{y-f(p)}{x-p}\implies y-f(p)=f'(p)(x-p)$$ as the tangent to the curve at the point $(p,f(p))$.
Let $dx=x-p$ and let $dy=y-f(p)$. Then the tangent line becomes $$dy=f'(p)dx$$ and this is the map of the tangent space of $\mathbb R$ at $x=p$ to the tangent space of $\mathbb R$ at $y=f(p)$.
The elements of the tangent space at $x=p$ are the $dx$s and are the changes we can make in any direction in $\mathbb R$. $\mathbb R$ is one dimensional so there is only one direction of change and that is along the $x$ axis.
The same analysis can be done with a function $g:\mathbb R^n\to\mathbb R$. Consider a point $q=(q_1,\dots,q_n)\in\mathbb R^n$.
We have $y=g(x_1,\dots\,x_n)$ and $$dy=\frac{\partial y}{\partial x_1}(q_1,\dots,q_n)dx_1+\dots +\frac{\partial y}{\partial x_n}(q_1,\dots,q_n)dx_n=\nabla g(q_1,\dots,q_n)\cdot d\vec x=\nabla g(q_1,\dots,q_n)\begin{pmatrix}dx_1\\ \vdots\\ dx_n\end{pmatrix}$$.
The derivative is represented by the gradient which is a linear functional from the tangent space of $\mathbb R^n$ at $q=(q_1,\dots,q_n)$ to the tangent space of $\mathbb R$ at $y=g(q_1,\dots,q_n)$. Note that $dx_i$ is the change along the $x_i$ axis in $\mathbb R^n$.