distinguish $\neg P\implies (Q\land\neg R)$ and $(\neg P\land Q)\implies\neg R $ in application The function $f:\mathbb{R}\to\mathbb{R} $ is continuous at $a\in\mathbb{R}$
if and only if
if $\displaystyle\lim_{n\to\infty} x_n=a$, then $\displaystyle\lim_{n\to\infty}f(x_n)=f(a).$
Denoting $P$  as the $\epsilon-\delta$ continuity definition, $Q$ as $\displaystyle\lim_{n\to\infty} x_n=a$, and $R$ as $\displaystyle\lim_{n\to\infty}f(x_n)=f(a),$ we have $$P\implies (Q\implies R)\equiv(P\land Q)\implies R$$ and $$(Q\implies R)\implies P\equiv \neg P\implies \neg(Q\implies R)\equiv\neg P\implies (Q\land \neg R).$$
However, the capture seems to be proving $$(\neg P\land Q)\implies \neg R.$$

How to distinguish $$(\neg P\land Q)\implies \neg R$$  from $$\neg P\implies (Q\land \neg R)$$ in this application?
Add-on:
I know that $\neg P\implies (Q\land \neg R)$ is the true formulation and know how to obtain it. What really bugs me is the capture uses $Q, \lim_{n\to\infty}x_n=a$, not arrived this.
 A: 
What really bugs me is the capture uses $Q, \lim_{n\to\infty}x_n=a$, not arrived this.

No, the screencap is assuming only that $P$ is false, and proving this:
$$\lnot P\implies \text{there is some }\langle x_n:n{\in}\mathbb N\rangle \text{ such that }\big(Q\land\lnot R\big);\tag1$$
equivalently (by contrapositive), it is proving that $$\Big[\text{for each }\langle x_n:n{\in}\mathbb N\rangle, \big(Q\implies R\big)\Big]\implies P.$$

the argument seems having Q in antecedent. If not, what will $$(¬P∧Q)⟹¬R\tag2$$ looks like in plain language?

Statement $(2)$ translates as “Suppose that $P$ is false but $Q$ true; we shall now show that $R$ is false” (interpreted: “Suppose that $f$ is discontinuous at $a$ and that $\langle x_n:n{\in}\mathbb N\rangle$ converges to $a;$ we shall now show that $f(x_n)$ does not converge to $f(a)”).$
The problem with this suggestion is that the sequence $\langle x_n:n{\in}\mathbb N\rangle$ in it is hanging freely: while the original theorem/definition that started your post was referring to every possible sequence $\langle x_n:n{\in}\mathbb N\rangle$ that converges to $a,$ and the screencap is referring to some sequence $\langle x_n:n{\in}\mathbb N\rangle$ coverging to $a,$ in statement $(2),$ a particular sequence appears to be referred to, yet it is unidentified.

The function $f:\mathbb{R}\to\mathbb{R} $ is continuous at $a\in\mathbb{R}$
if and only if
if $\displaystyle\lim_{n\to\infty} x_n=a$, then $\displaystyle\lim_{n\to\infty}f(x_n)=f(a).$

To be clear: you've probably transcribed the theorem/definition incorrectly: $$\text{for each }\langle x_n:n{\in}\mathbb N\rangle$$ ought to be inserted at the beginning of its third line.
A: No, the logical form of the argument is correct and does what it is supposed to do.
First, I try to adapt the logical form of the argument to your formalization. Then, I show that your formalization is missing something important, the quantifiers, to give a full account for the logical structure of the proof argument.

The proof you cited shows that if the condition about limits is fulfilled (the $Q \implies R$ part of your formalization, i.e. for every sequence $\{x_n\}_{n\in \mathbb{N}}$, if $\lim_{n \to \infty} x_n = a$ then $\lim_{n \to \infty} f(x_n) = f(a)$), then $f$ is continuous in $a$ (the $P$ part of your formalization). The proof uses the technique called "by contrapositive": it supposes that the function $f$ is not continuous at $a$ (i.e. it assumes $\lnot P$) and then it shows that the condition about limits is not fulfilled, i.e. it shows that $Q \land \lnot R$ (which is equivalent to $\lnot (Q \implies R)$).
Indeed, the proof argument shows that there exists a sequence  $\{x_n\}_{n\in \mathbb{N}}$ such that $\lim_{n \to \infty} x_n = a$ (i.e. $Q$ holds) but it is not true that $\lim_{n \to \infty} f(x_n) = f(a)$ (i.e. $\lnot R$ holds).
Note that this part of the proof you quoted is maybe badly worded and a bit confusing. You do not suppose to have a sequence $\{x_n\}_{n \in \mathbb{N}}$ such that $\lim_{n \to \infty} x_n = a$ (the phrase "Also $\lim_{n \to \infty} c_n = c$" is misleading there, because the sequence $\{c_n\}_{n \in \mathbb{N}}$ is not defined yet). Instead, you construct a sequence $\{x_n\}_{n \in \mathbb{N}}$ for which it turns out that  $\lim_{n \to \infty} x_n = a$.
How?
Since $f$ is not continuous in $a$ by hypothesis, we know that
$$\text{there is $\varepsilon \!>\! 0$ such that, for all $\delta \!>\! 0$, $|x - a| \!<\! \delta$ and $|f(x) - f(a)| \!\geq\! \varepsilon$ for some $x$}\tag{*}$$
For every $n \in \mathbb{N}^{>0}$, let $x_n$ be the real number such that $|x_n - a| < \frac{1}{n}$ and $|f(x_n) - f(a)| \geq \varepsilon$.
Such a $x_n$ exists thanks to $(\text{*})$, by taking $\delta = \frac{1}{n}$.
From that definition of $\{x_n\}_{n \in \mathbb{N}}$, it is immediate to see that $\lim_{n \to \infty} x_n = a$ and that $\lim_{n \to \infty} f(x_n) \neq f(a)$.
See also the proof quoted here to have an explanation of the argument that, in my opinion, is better than the one you quoted in the OP.

What is missing in your formalization of the logical form of the argument? The universal quantification about sequences. I guess this can partially be the source of your misunderstanding.
Indeed, a more faithful formalization of the proof argument would be of the form: we assume $\lnot P$ and we show that this implies that $\lnot \forall s (Q(s) \implies R(s))$, which amounts to show that $\exists s (Q(s) \land \lnot R(s))$.
Here, by $Q(s)$ I mean that the sequence $s$ converges to $a$, and by $R(s)$ I mean that the limit of $f$ applied to each element of the sequence $s$ is $f(a)$.
Said differently, in order to prove that the function $f$ is continuous in $a$ (i.e. to prove that $P$ holds), it is sufficient to prove that for every sequence $s$, if $Q(s)$ then $R(s)$. By contrapositive, it amounts to show that  if $\lnot P$ does not hold, then there exists some sequence $s$ such that $Q(s)$ and $\lnot R(s)$ hold.
