Dear Ritwik -- This is not a precise reference for the result you state. However, standard dimension results, i.e., Exercise II.3.22, p. 95 of Hartshorne's *Algebraic Geometry*, lead to a proof very quickly (and I would consider this as a standard application of dimension theory). Let $\overline{\mathcal{C}}\subset \overline{\mathcal{A}}\times \mathbb{P}^2$ be the universal curve over $\overline{\mathcal{A}}$, i.e., a (geometric) point of $\overline{\mathcal{C}}$ parameterizes a pair $([f],p)$ of a curve $[C]=[Z(f)]$ in $\overline{\mathcal{A}}$ and a point $p\in C$, possibly $p=[1,0,0]$. Note that the projection $$\pi:\overline{\mathcal{C}}\to \overline{\mathcal{A}}$$ is a flat morphism of relative dimension $1$. By the way, the other projection, $$\text{ev}:\overline{\mathcal{C}}\to \mathbb{P}^2$$ is certainly **not** flat, since the fiber over $[1,0,0]$ has larger dimension than the fiber over any other point $p$.

Also define $\mathcal{C}$ to be the open subset of $\overline{\mathcal{C}}$ which is the intersection of the open subsets $\pi^{-1}(\mathcal{A})$ and the smooth locus of the morphism $\pi$, i.e., in your terminology the open locus where $\nabla f|_p\neq 0$. Concretely, $\mathcal{C}$ parameterizes pairs $([C],p)$ where $C$ has a strict node at $[1,0,0]$ and where $p$ is a smooth point of $C$.

Now consider the $k$-fold fiber product $$\overline{\mathcal{C}}^k := \overline{\mathcal{C}}\times_{\overline{\mathcal{A}}} \dots \times_{\overline{\mathcal{A}}} \overline{\mathcal{C}}.$$ This parameterizes data $([C],p_1,\dots,p_k)$ of a curve $[C]=[Z(f)]$ in $\overline{\mathcal{A}}$ and $k$ points $p_1,\dots,p_k$ in $C$, not necessarily smooth and not necessarily distinct. Similarly, define $\mathcal{C}^k$ to be the $k$-fold fiber product $\mathcal{C}\times_{\mathcal{A}}\times \dots \times_{\mathcal{A}} \mathcal{C}$. This is an open subscheme of $\overline{\mathcal{C}}^k$. It parameterizes data $([C],p_1,\dots,p_k)$ such that $[C]$ is in $\mathcal{A}$ and such that $p_1,\dots,p_k$ are smooth points of $C$ (but possibly coincident -- this will be irrelevant soon).

There are a couple of observations. First, as a $k$-fold fiber product of flat morphisms of relative dimension $1$, the projection morphism, $$\pi_k:\overline{\mathcal{C}}^k \to \overline{\mathcal{A}},$$ is also flat of relative dimension $k$; thus $\overline{\mathcal{C}}^k$ has pure dimension $2k$. Similarly, since the projection $\mathcal{C}\to \mathcal{A}$ is smooth of relative dimension $1$, also the projection, $$\pi:\mathcal{C}^k\to \mathcal{A},$$ is smooth of relative dimension $k$; thus $\mathcal{C}^k$ is smooth of pure dimension $2k$. Second, the open subscheme $\mathcal{C}^k$ is dense in $\overline{\mathcal{C}}^k$, i.e., this open subscheme intersects every irreducible component (in fact there is only one irreducible component). To prove this, first note that every irreducible component dominates $\overline{\mathcal{A}}$ since $\pi$ is flat. Thus it suffices to prove that for a general point $[C]$ of $\overline{\mathcal{A}}$, the fiber of $\pi_k$ over $[C]$ intersects $\mathcal{C}^k$ in a dense open. As discussed in the previous post, $C$ is smooth away from $[1,0,0]$. The fiber of $\pi_k$ over $[C]$ is canonically isomorphic to the $k$-fold product $C^k := C\times \dots \times C$ (of course all *absolute* fiber products are relative to the Spec of our ground field). And the intersection with $\mathcal{C}^k$ is the fiber product $(C\setminus \{[1,0,0]\})^k$, which is clearly dense since the complement is the closed subset of strictly smaller dimension $(k-1)$ which is the union of the $k$ subvarieties of the form $$C\times \dots \times C \times \{[1,0,0]\} \times C \times \dots \times C.$$ Because $\mathcal{C}^k$ is dense in $\overline{\mathcal{C}}^k$, the complement $D$ is a proper subvariety of dimension $\leq 2k-1$.

Finally we are ready to set up the application of dimension theory. Consider the $k$-fold product of the morphism $\text{ev}$, $$\text{ev}:\overline{\mathcal{C}}^k \to (\mathbb{P}^2)^k, \ ([C],p_1,\dots,p_k)\mapsto (p_1,\dots,p_k).$$ For the smooth open $\mathcal{C}^k$ of pure dimension $2k$, Exercise II.3.22(e) states that there exists a dense open subset $W\subset (\mathbb{P}^2)^k$ such that for every point $(p_1,\dots,p_k)$ of $W$, the fiber $g^{-1}(p_1,\dots,p_k)\cap \mathcal{C}^k$ has the expected dimension $0$ (or it is empty). Next, we can write the closed complement $D$ of $\mathcal{C}^k$ as a union of locally closed subsets, $$D=D_1\sqcup \dots \sqcup D_r,$$ such that each $D_i$ is smooth and irreducible. As a locally closed subset of $D$, $\text{dim}(D_i) \leq \text{dim}(D) < 2k$. Since $(\mathbb{P}^2)^k$ has dimension $2k$, none of the morphisms $D_i \to (\mathbb{P}^2)^k$ are dominant, i.e., the image of each is contained in a proper, closed subset $E_i\subset (\mathbb{P}^2)^k$. Thus, up to replacing $W$ by the dense open subset $W\setminus (E_1\cup \dots E_r$, we conclude that for every point $(p_1,\dots,p_k)$ of $W$, the fiber $f^{-1}(p_1,\dots,p_k)\cap D_i$ is either empty or has the expected dimension $-1$. Since a variety cannot have negative dimension, Kleiman-Bertini precisely says that the fiber is empty. In other words, the fiber $f^{-1}(p_1,\dots,p_k)$ is disjoint from the closed subset $D$. Stated yet again, the fiber $f^{-1}(p_1,\dots,p_k)$ is completely contained in $\mathcal{C}^k$. Since $\pi(f^{-1}(p_1,\dots,p_k))$ equals $\overline{A}\cap H_{p_1}\cap \dots \cap H_{p_k}$, this gives your result.

In a manuscript, I believe that most algebraic geometry readers would be able to supply the argument above if you simply state that your result follows from dimension arguments.

**Edit.** In the first version of the answer, I used the Kleiman-Bertini theorem. This is valid (and is my own first instinct as an algebraic geometer). However, since there are issues with Kleiman-Bertini in positive characteristic, I feel it is better to phrase the argument in terms of dimension theory instead.