This question is motivated by this MO question. An argument that I would like to explain now I can't work out.
Let $C$ be an irrecucible, reduced curve (=1D proper scheme) over an alg closed field $k$ and $L$ line bundle/invertible sheaf that provides a morphism $f: C \to \mathbb{P}^2$.
Recall that a morphism to $\mathbb{P}^2$ is equivalent to the data $(L, s_0, s_1, s_2)$ i.e. of a line bundle $L$ and $3$ global sections $s_i \in H^0(C,L)$. These are related to $f$ as follows: $f$ induces a map of $k$-vector spaces $kX_0 \oplus kX_1 \oplus kX_2=H^0(\mathbb{P}^2, \mathcal{O}(1)_{\mathbb{P}^2}) \to H^0(C, L), X_i \mapsto f^*X_i =s_i$. Thus $s_i$ are nothing but the pullbacks of canonical generators of $H^0(\mathbb{P}^2, \mathcal{O}(1)_{\mathbb{P}^2})$ by $f$: the $X_i$.We write suggestively $f= [s_0: s_1:s_2]$ since multiplying the vector $(s_0,s_1,s_2)$ by a nonzero constant not change $f$.
Obviuosly, we obtain intrinsical factorization $f: C \to C' \subset \mathbb{P}^2$ with $C' = \overline{f(C)}$. Denote the inclusion $i:C' \subset \mathbb{P}^2$. As before the inclusion $i$ is determined by three glocal sections $f_0,f_1,f_2$ of line bundle $i^*\mathcal{O}_{\mathbb{P}^2}(1)= \mathcal{O}_{\mathbb{P}^2}(1) \vert_{C'}$ (last one is more suggestively notion).
By functoriality of the constrution the map $H^0(\mathbb{P}^2, \mathcal{O}(1)_{\mathbb{P}^2}) \to H^0(C, L)$ factorizes through $H^0(C', \, \mathcal{O}_{\mathbb{P}^2}(1)|_{C'})$ via $X_i \mapsto i^*X_i=f_i \mapsto s_i$. Thus we obtain a pullback map (??? compare with argument from MO question) $$f^*:H^0(C', \, \mathcal{O}_{\mathbb{P}^2}(1)|_{C'}) \to H^0(C, L)$$
Question 1: why this map of $k$-spaces is well defined and injective if we know that $s_0,s_1,s_2$ are linearly independent? The construction tells what happens with $f_i$ but I don't see the reason why $f$ determines what happens in the whole space $H^0(C', \, \mathcal{O}_{\mathbb{P}^2}(1)|_{C'})$. Equivalently, why is $i^*:H^0(\mathbb{P}^2, \mathcal{O}(1)_{\mathbb{P}^2}) \to H^0(C', \, \mathcal{O}_{\mathbb{P}^2}(1)|_{C'})$ surjective?
Motivation: in the MO question above I asked for the reason of $\dim_k H^0(C,L^m) \ge \dim_k H^0(C', \mathcal{O}_{\mathbb{P}^2}(m) \vert _{C'})$. That's exacly the case $m=1$.
Question 2: How from case $m=1$ in Q1 it can be derived that the twisted version $H^0(C', \, \mathcal{O}_{\mathbb{P}^2}(m)|_{C'}) \to H^0(C, L^m)$ is still injective?