First let me simplify the clutter of notation a bit; set $S:=V\otimes_RR[t]$ and $E:=\operatorname{End}_{R[t]}(S)$.
Throughout this answers I will view elements of $R[f,t]$ as $R[t]$-linear endomorphisms of $S$, i.e. as elements of $E$. As for exterior powers; for the proof only the case $k=1$ is relevant, so I won't bother with them at all.
The idea of the proof is to show that the endomorphism $\chi_f\in E$ vanishes on the quotient $S/(f-t)S$, and then to show that $S/(f-t)S\cong V$ as $R$-modules. The main ingredient is showing that $f-t\in E$ commutes with its adjugate. This relies on the fact that $\chi_f$ is not a zero divisor in $R[t]$.
The proof is a lot of commutative algebra, I have assumed everything in Atiyah-Maconald. If any part is unclear, let me know.
Step 1: The characteristic polynomial is not a zero divisor in $R[t]$.
The characteristic polynomial $\chi_f$ of $f-t\in R[f,t]$ is the determinant of the $R[t]$-linear map $f-t\in E$. Note that $\chi_f\in R[t]$ is not a zero divisor because $f-t\in E$ is injective, because its leading coefficient as a polynomial in $t$, i.e. as an element of $(R[f])[t]$, is a unit.
Step 2: The endomorphism $f-t\in E$ commutes with its adjugate w.r.t. the given pairing.
The adjugate of $f-t\in E$ with respect to the given perfect pairing is the unique $F\in E$ such that
$$F\cdot(f-t)=\chi_f\cdot1_S.\tag{1}$$
Because $\chi_f\in R[t]$ is not a zero divisor, localizing at $\chi_f$ yields an injection $R[t]\ \longrightarrow\ R[t]_{\chi_f}$. Because $V$ is a finitely generated free $R$-module, this in turn yields injections
$$S\ \longrightarrow\ S_{\chi_f}
\qquad\text{ and }\qquad
E\ \longrightarrow\ E_{\chi_f}.$$
By construction $\chi_f$ is a unit in $E_{\chi_f}$ and hence $(1)$ shows that also $f-t$ is a unit in $E_{\chi_f}$, so
$$F=\chi_f\cdot(f-t)^{-1},$$
in $E_{\chi_f}$. This shows that $F$ and $f-t$ commute in $E_{\chi_f}$, because both are $R[t]$-linear and $\chi_f\in R[t]$. Because $E_{\chi_f}$ contains $E$ as a subring, they also commute in $E$.
Step 3: On the quotient module $S/(f-t)S$ we have $\chi_f(f)=0$.
Because $F$ and $f-t$ commute, for all $(f-t)s\in(f-t)S$ we have
$$F((f-t)s)=(f-t)F(s)\in(f-t)S,$$
so $F$ maps the $S$-submodule $(f-t)S\subset S$ into itself. This means $F$ descends to an $R[t]$-linear map
$$S/(f-t)S\ \longrightarrow\ S/(f-t)S.$$
In this quotient $f-t$ is identically zero, so identity $(1)$ shows that on the quotient
$$F\cdot0=\chi_f\cdot1_{S/(f-t)S},$$
and so $\chi_f$ is identically zero on $S/(f-t)S$, where of course $\chi_f(t)=\chi_f(f)$ on the quotient.
Step 4: Also $\chi_f(f)=0$ on $V$.
Because $\chi_f(f)=0$ on $S/(f-t)S$ and the composition
$$V\ \longrightarrow\ S\ \longrightarrow\ S/(f-t)S,$$
is an isomorphism of $R[f]$-modules, it follows that $\chi_f(f)=0$ on $V$.
Best Answer
I'm going to denote adjoints by $*$ rather than $\dagger_g$, for notational simplicity.
First answer:
Yes if $g$ is a perfect pairing, then adjoints always exist and are unique.
Let's exploit the tensor-hom adjunction and let $g_V: W\to \newcommand\Hom{\operatorname{Hom}}\Hom(V,L)$ be the obvious map. Now for any $f$, we can consider the map $g_f : W\to \Hom(V,L)$ defined by $g_f(w) = g_V(w)\circ f$. Then in order for an adjoint to exist, we must be able to solve the equation $$ g_Vf^* = g_f.$$
Therefore if $g_V$ is an isomorphism (i.e. if $g$ is a perfect pairing), there is a unique $f^*$ satisfying the equation, $f^*=g_V^{-1}g_f$.
Second answer:
Let's generalize slightly. When can we solve $g_Vf^*=g_f$? Consider the following diagram $$\require{AMScd} \begin{CD} W @>g_f>> \Hom(V,L) \\ @Vf^*VV @| \\ W @>>g_V> \Hom(V,L) \end{CD} $$
Well, one answer to when we can find such a $f^*$ is if $g_V$ is surjective and $W$ is projective. In this case $f^*$ won't be unique. In fact, this is roughly the most general we can get, though.
To generalize this slightly, observe that the image of $g_f$ had better be a subset of the image of $g_V$, otherwise there is no way we can solve it. However if the image of $g_f$ is a subset of the image of $g_V$, then we can replace $\Hom(V,L)$ with $\newcommand\im{\operatorname{im}}\im g_V$, so that $g_V$ is now surjective to its image. Then as long as $W$ is projective, we can lift $g_f$ along $g_V$ to find $f^*$.
My final version of the second answer:
As long as $W$ is projective, and for every $w$, there exists $w'$ so that $g_V(w)\circ f = g_V(w')$, then there exists a (possibly not unique) "adjoint" $f^*$ solving $g_Vf^* = g_f$.