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:
$$
r(D) = (g-2) - (d'-1 - r(D')) = (g-1) - (d' - r(D'))
$$
Riemann Roch tells you that for divisors of sufficiently large degree, the global sections of $O_Y(D)$ have dimension $\deg(D)-g_Y+1$, and this lets you build sections with desired zero sets. In your case, consider a large positive divisor $D$ of support disjoint from your point $P$, and consider the sections of $O_Y(D)$ and $O_Y(D+P)$, viewing these as subsheaves of the fraction field $K(Y)$. As subsheaves of $K(Y)$ these are given by $$L(D)(U)=\{f\in K(X)|\text{div}(f)+D|_U\geq 0\}$$
By our sufficiently large hypothesis, there exists a global section of $L(D+P)$ that doesnt arise from $L(D)$, which after translating through the definitions gives a function $f$ in $K(X)$ which has an order $1$ pole at $P$, and all other poles contained in the support $D$, and zero set also disjoint from $P$. So by construction we have the divisor $Z=P+\text{div}(f)$ disjoint from $P$.
The normality here is critical, in that we needed our curve to be nonsingular to use Riemann Roch, and normal=nonsingular in dimension $1$, so by normalising, we resolved the singularities of $X$.
Best Answer
If you work in the function field case, i.e. replace the number field $K$ and its places $v$ by a finite extension of $k[x]$ for some finite field $k$, and work with its places, then this statement becomes the Riemann--Roch theorem.
The appearance of $1-g$ in the RR thm., and the role of the canonical bundle in forming the correct kind of dual, will here be absorbed into the the definition of the self-dual measure on the adeles.
To be a little more precise, you should imagine that your line bundle is of the form $\mathcal L(D)$ for some divisor $D$ (it always is, after all!); then $a$ will play the role of $D$ (or maybe $-D$). And you should take $f$ to the characteristic function of the integral adeles. Then one side of the equality will count rational functions $\xi$ for which $\xi a^{-1}$ has no demoninators (so global sections of $\mathcal L(D)$), and the other side will count rational functions $\xi$ such, roughly, $\xi a^{-1}$ is integral, which is the global sections of $\mathcal L(-D)$; except that $f$ is not quite self-dual. In the number field case the different comes in, and in the function field case we are considering here, the canonical bundle will come in (as well as a factor related to $1-g$).
Finally, to get the familiar statement about dimensions, take log of both sides and divide by $\log q$ (where $q = |k|$).
(You can check then that the $|a|^{-1}$ on the LHS, after taking logs, gives the $\deg D$ term.)