Is this case of weighted 2SAT NP-complete? Weighted 2SAT asks if it is possible to satisfy the formula with at most $k$ variables set as positive/negative. Trivially, every instance must be in 2CNF. It is known to be $\mathsf{NP}$-complete.

We have following additional restriction: each variable appears twice as positive and once as negative or vice versa.

Example of instance:

$(x\lor y)\land(x\lor z)\land(\overline x\lor t)\land(\overline y\lor z)\land(\overline y\lor \overline t)\land(\overline z\lor \overline t)$

Is this weighted 2SAT variant $\mathsf{NP}$-complete?

Of course, if vertex cover where each vertex has only 2 edges is $\mathsf{NP}$-hard, this also must be $\mathsf{NP}$-hard. So, if such result is known, then it also can be an answer. Ah, well, this does not help.
You can express the predicate "$x = y$" using one occurrence of each polarity:
$$
(x \lor \lnot y) \land (\lnot x \lor y).
$$
Consider now an instance of weighted 2SAT, in which each variable appears at most $M$ times. Duplicate each variable $M$ times, and enforce that all copies are the same using the gadget above. Replace each occurrence of each variable by a distinct copy of the variable. If the original instance asks for an assignment with at most $k$ positive variables, ask for at most $Mk$ positive variables. We obtain an instance of your problem which is equivalent to the original problem.

This shows that your problem is also NP-complete.