The case $d > 1$. Since we have a compatible system of maps $p^*(I^q) \to \mathcal{O}_X(q)$ for $q \geq 0$ there are canonical maps $p^*(I^q/I^{q + \nu}) \to \mathcal{O}_{Y_\nu}(q)$ for $\nu \geq 0$. Using this and pulling back $\varphi$ we obtain a map $$ \chi : p^*M \longrightarrow \mathcal{O}_{Y_n}(n) $$ such that the composition $M \to H^0(X, p^*M) \to H^0(X, \mathcal{O}_{Y_n}(n))$ is the given homomorphism $\varphi$ combined with the map $I^n/I^{2n} \to H^0(X, \mathcal{O}_{Y_n}(n))$. Since $\mathcal{O}_{Y_n}(n)$ is invertible on $Y_n$ the linear map $\chi$ determines a section $$ \sigma \in \Gamma(X, (p^*M)^\vee(n)) $$ with notation as in Remark \ref{remark-duals}. The discussion in Remark \ref{remark-duals} shows the cokernel and kernel of $can : p^*(M^\vee) \to (p^*M)^\vee$ are scheme theoretically supported on $Y_c$. By Lemma \ref{lemma-annihilated} the map $(p^*M)^\vee(n) \to (p^*M)^\vee(n - 2c)$ factors through $p^*(M^\vee)(n - 2c)$; small detail omitted. Hence the image of $\sigma$ in $\Gamma(X, (p^*M)^\vee(n - 2c))$ comes from an element $$ \sigma' \in \Gamma(X, p^*(M^\vee)(n - 2c)) $$ By Lemma \ref{lemma-vanishing-coh-almost-projective} part (5), the fact that $M^\vee$ is an $(A, n, c)$-module by More on Algebra, Lemma \ref{more-algebra-lemma-dual-near-projective}, and the fact that $n \geq q_0 + (1 + d)c$ so $n - 2c \geq q_0 + (d - 1)c$ we see that the image of $\sigma'$ in $H^0(X, p^*M^\vee(n - (1 + d)c))$ is the image of an element $\tau$ in $I^{n - (1 + d)c}M^\vee$. Write $\tau = \sum a_i \tau_i$ with $\tau_i \in I^{n - (2 + d)c}M^\vee$; this makes sense as $n - (2 + d)c \geq 0$. Then $\tau_i$ determines a homomorphism of modules $\psi_i : M \to I^{n - (2 + d)c}/I^{2n - (2 + d)c}$ using the evaluation map $M \otimes M^\vee \to A/I^n$. \medskip\noindent Let us prove that this works\footnote{We hope some reader will suggest a less dirty proof of this fact.}. Pick $z \in M$ and let us show that $\varphi(z)$ and $\sum a_i \psi_i(z)$ have the same image in $I^{n - (1 + d)c}/I^{2n - (1 + d)c}$. First, the element $z$ determines a map $p^*z : \mathcal{O}_{Y_n} \to p^*M$ whose composition with $\chi$ is equal to the map $\mathcal{O}_{Y_n} \to \mathcal{O}_{Y_n}(n)$ corresponding to $\varphi(z)$ via the map $I^n/I^{2n} \to \Gamma(\mathcal{O}_{Y_n}(n))$. Next $z$ and $p^*z$ determine evaluation maps $e_z : M^\vee \to A/I^n$ and $e_{p^*z} : (p^*M)^\vee \to \mathcal{O}_{Y_n}$. Since $\chi(p^*z)$ is the section corresponding to $\varphi(z)$ we see that $e_{p^*z}(\sigma)$ is the section corresponding to $\varphi(z)$. Here and below we abuse notation: for a map $a : \mathcal{F} \to \mathcal{G}$ of modules on $X$ we also denote $a : \mathcal{F}(t) \to \mathcal{F}(t)$ the corresponding map of twisted modules. The diagram $$ \xymatrix{ p^*(M^\vee) \ar[d]_{can} \ar[r]_{p^*e_z} & \mathcal{O}_{Y_n} \ar@{=}[d] \\ (p^*M)^\vee \ar[r]^{e_{p^*z}} & \mathcal{O}_{Y_n} } $$ commutes by functoriality of the construction $can$. Hence $(p^*e_z)(\sigma')$ in $\Gamma(Y_n, \mathcal{O}_{Y_n}(n - 2c))$ is the section corresponding to the image of $\varphi(z)$ in $I^{n - 2c}/I^{2n - 2c}$. The next step is that $\sigma'$ maps to the image of $\sum a_i \tau_i$ in $H^0(X, p^*M^\vee(n - (1 + d)c))$. This implies that $(p^*e_z)(\sum a_i \tau_i) = \sum a_i p^*e_z(\tau_i)$ in $\Gamma(Y_n, \mathcal{O}_{Y_n}(n - (1 + d)c))$ is the section corresponding to the image of $\varphi(z)$ in $I^{n - (1 + d)c}/I^{2n - (1 + d)c}$. Recall that $\psi_i$ is defined from $\tau_i$ using an evaluation map. Hence if we denote $$ \chi_i : p^*M \longrightarrow \mathcal{O}_{Y_n}(n - (2 + d)c) $$ the map we get from $\psi_i$, then we see by the same reasoning as above that the section corresponding to $\psi_i(z)$ is $\chi_i(p^*z) = e_{p^*z}(\chi_i) = p^*e_z(\tau_i)$. Hence we conclude that the image of $\varphi(z)$ in $\Gamma(Y_n, \mathcal{O}_{Y_n}(n - (1 + d)c))$ is equal to the image of $\sum a_i\psi_i(z)$. Since $n - (1 + d)c \geq q_0$ we have $\Gamma(Y_n, \mathcal{O}_{Y_n}(n - (1 + d)c)) = I^{n - (1 + d)c}/I^{2n - (1 + d)c}$ by Lemma \ref{lemma-bound-q-and-d} and we conclude the desired compatibility is true. \medskip\noindent The case $d = 1$. Here we argue as above that we get $$ \chi : p^*M \longrightarrow \mathcal{O}_{Y_n}(n),\quad \sigma \in \Gamma(X, (p^*M)^\vee(n)),\quad \sigma' \in \Gamma(X, p^*(M^\vee)(n - 2c)), $$ and then we use Lemma \ref{lemma-vanishing-coh-almost-projective} part (4) to see that $\sigma'$ is the image of some element $\tau \in I^{n - 2c}M^\vee$. The rest of the argument is the same. \medskip\noindent The case $d = 0$. Argument is exactly the same as in the case $d = 1$.