The answer to your first question is yes. As for the second question I would make a minor adjustment but the idea is correct. Please be patient with me as I set things up first.
Notation
Let $\omega_C$ be the sheaf of holomorphic differential forms on $C$. For any vector space $V$ I will denote by $|V|$ the projective space of lines in $V$ and by $\mathbb{P}(V)$ the projective space of codimension 1 planes in $V$, henceforth referred to as hyperplanes.
Throughout I will let $V = H^0(C,\omega_C)$ stand for the global holomorphic differentials on $C$. For a divisor $D$ on $C$ let $V(-D) = H^0(C,\omega_C(-D))$ which is to be viewed as a subspace of $V$ using the natural injection $\omega_C(-D) \hookrightarrow \omega_C$.
Let me write the map $\varphi : C \to \mathbb{P}^{g-1}$ with the proper notation, because as things stand the notation suggests that generators for $V$ have been chosen. The preferred version, as you put it, can be written as $\varphi: C \to \mathbb{P}(V)$ where
$$
p \mapsto V_p := \mathrm{ker}(V \to \omega_C \to \omega_C|_p).
$$
It is clear that $V(-p) \subset V_p$ but Riemann-Roch says that the dimensions match so we get $V(-p) = V_p$.
Basic observations
Given $q \in \mathbb{P}(V)$ we get a hyperplane $H_q \subset |V|$ by duality. Here, $H_q$ parametrizes hyperplanes in $\mathbb{P}(V)$ containing $q$. Given two points $q_1, q_2$ the line between $q_1$ and $q_2$ can be viewed as the intersection of all hyperplanes containing $q_i$'s. This is the dual of $H_{q_1}\cap H_{q_2}$. And similarly for more points. So far this is linear algebra, let us apply it to our specific setting.
Our description of the map $\varphi$ ensures that $H_{\varphi(p)} = |V(-p)|$. Given $p_1, p_2 \in C$ the line between $\varphi(p_1)$ and $\varphi(p_2)$ has dual $|V(-p_1)|\cap|V(-p_2)|$ which is easily seen to be $|V(-p_1-p_2)|$ (Check!). And if $D$ is reduced, this argument is sufficient to conclude that $|V(-D)| \subset |V|$ parametrizes the hyperplanes through $\varphi(D)$.
It is possible, but difficult, to describe the locus of hyperplanes having high order contact at a point $p \in C$ using just geometry. So we go back to doing a little more tautology.
Advanced tautology
If you give me any divisor $D \in |\omega_C| = |V|$ then this gives a line in $H^0(C,\omega_C)$. Pick a section $\sigma$ generating this line. By the standard relations $D = (\sigma)_0$. Now this point $D$ has a dual hyperplane $W_D \subset \mathbb{P}(V)$ consisting of all hyperplanes in $|V|$ containing $D$. I claim that $W_D \cdot \varphi(C) = D$.
Indeed, $W_D$ corresponds to a divisor in the linear system of the tautological line bundle $\mathcal{O}(1)$ which by construction has global sections canonically isomorphic to $V$. Furthermore, the divisor $W_D$ corresponds precisely to the (line generated by) $\sigma \in V$. Now the intersection of $W_D$ with $C$ can be obtained by pulling back $(\mathcal{O}(1),\sigma)$ to $C$. However, the pullback of $\mathcal{O}(1)$ is of course (canonically isomorphic to) $\omega_C$ and the section $\sigma$ maps to $\sigma$ again by canonical identifications. This proves the desired statement.
Wrapping up
So far what we have been doing was largely exploratory. Now let's tackle your question head on. The inclusion $V(-D) \hookrightarrow V$ identifies sections of $\omega_C(-D)$ with sections of $\omega_C$ that vanish on $D$ (this follows from the exact sequence obtained from $\omega_C(-D) \hookrightarrow \omega_C$). Therefore if we take $\sigma \in V(-D)$ then the corresponding hyperplane $H \subset \mathbb{P}(V)$ satisfies $H \cdot C \ge D$. Conversely, any hyperplane that satisfies $H \cdot C \ge D$ corresponds to (the line generated by) a section of $\omega_C$ vanishing on $D$, hence to a section of $\omega_C(-D)$. This answers your first question.
As for the second question, if $d' = \mathrm{deg} D'$ and $d' < g$ then we expect $\varphi(D')$ to span a $d'-1$ dimensional projective subspace. Then your calculation shows that $r(D')$ in fact measures the failure of our expectation. More precisely, $r(D') = d'-1 - \mathrm{dim}(\mathrm{span}(\varphi(D')))$. The space of hyperplanes in $\mathbb{P}^{g-1}$ passing through a $k$-dimensional projective space has dimension $(g-2) - k.$ Then putting these together we get:
$$
r(D) = (g-2) - (d'-1 - r(D')) = (g-1) - (d' - r(D'))
$$
If $D$ is non-negative ok, if not just change $D$ by $D'=div(f)+D$ where $f\in O_D$, then $l(D)=l(D')\leq 1+deg(D')=1+deg(D)$.
rmk$_1$: If $D$ and $D'$ are equivalent divisors on Riemann Surfaces then $O_D(X)\simeq O_{D'}(X)$ given by multiplication of a meromorphic function on $X$.
rmk$_2$: $div(f)$ is the divisor associated a meromorphic function, then $deg(div(f))=0$ (The degree of a holomorphic map is constant in the set of regular values).
Best Answer
The statement of Asymptotic Riemann Roch should be
Using Serre vanishing, we deduce a statement about $h^0$ as a consequence.
Your example shows that if $D$ is not ample, this result may no longer hold. Indeed, $\pi^*D + E$ is not ample since it intersects negatively with $E$.