$(1) + (4)$ do not imply $q\lor r$, so undo line $(5)$ We do have that $(3) + (4)$ imply $p$. If $p \lor q$, and $\lnot q$, then $p$. Perhaps this can be line $(5)$.
Suggestion for the next step $(6)$: from $(3),$ along with the premise $\lnot q \rightarrow (u \land s)$, it follows by modus ponens that $u \land s$.
(7) Now, extract $s$ from $u \land s$.
(8) Then introduce the conjunction: $p \land s$. $p$ is from your new (5th) step, and $s$ from step (7).
Now we can use the premise $p\land s \implies t$ and $p \land s$ from step (8) to conclude by modus ponens that $t$, as desired
Key point: If you haven't used a premise, think of how it might help get you from what you have to what you need to establish. Always keep the goal or target proposition in your mind.
I'll give a summary "sketch" of one proof approach.
I'd suggest you start from the first premise:
First check what follows from $r\land \lnot s$ and then what follows when $q \land \lnot s$. (At least one of these must be true. Why?)
From each of the above, we can use conjunction elimination (or simplification) to obtain $\lnot s$. And so, we can use disjunction elimination to conclude $\lnot s$.
From $\lnot s$ and premise $(2)$, we can derive (infer), by modus ponens, $(p\land r)\rightarrow u$.
From $(p \land r) \rightarrow u$ together with premise $(3)$, we get $(p\land r) \rightarrow (s \land \lnot t)$
Now, we already derived $\lnot s.$ This means $\lnot s \lor t$ is true. Why?
But $\lnot s \lor t \equiv \lnot (s \land \lnot t).$
From $\lnot (s \land \lnot t),$ together with our derived $(p\land r )\rightarrow (s\land \lnot t)$, by modus ponens, we get $\not(p \land r)$
This means that either $\lnot p$, or $\lnot r$ is true.
If $\lnot p$ is true, we can build from this (addition) to get $\lnot p \lor q \equiv p\rightarrow q$, as desired.
If $\lnot r$ is true, then $r\land \lnot s$ is false (first alternative from premise $(1)$. Then it must follow that $q\land \lnot s$ is true. So it follows that $q$ is true (simplification). And given $q$, we have $\lnot p \lor q$ (addition), which is equivalent to $p \rightarrow q$, as desired.
Therefore, from $\lnot (p \land r)\equiv \lnot p \lor \lnot r$ we can conclude $(q\rightarrow q)$.
I'll leave this to you to write formally (including keeping proper track of temporary assumptions), with proper justifications.
Best Answer
I believe you'll need to use universal instantiation on $(2)$. As a universally quantified premise, you can instantiate with $\,a\,$, as you did with premise $(1)$, (since the domain of all the quantifiers is the same) then use modus tollens when you need it, applied to the instantiated statement and using the assumption $\lnot R$.
Also, as currently numbered, I think you need to justify resolution for line $(7)$ by citing $(3)$ and $(6)$ (I'm assuming you're referring to resolution to $\,P(a)\,$ given the statements $\,\left[P(a) \lor Q(a)\right]\;$ and $\left[P(a) \lor \lnot Q(a)\right].$
The key here is that you want your steps (currently numbered) $4 - 8$ indented (to designate a "sub-proof", where $(4)$ is stated as an "assumption", then $5$ follows from $(4)$ and the (soon to be) instantiated step resulting from $(2)$.
Line $(9)$ should still be a statement about the constant $a$, not $x$ as written, citing sub-proof, $(4 - 8)$ (by assuming (4), we obtain (8)), i.e. $(4)\rightarrow (8)$ given subproof $4-8$.
From $9$ to $10$ then, you are correct, as is your justification of universal generalization.