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 $V$.
My previous answer made a false claim -- that we wanted to view $M$ as an $M_{n\times n}(R[x])$-module. In fact, this will not work in general: while given an endomorphism $\phi\in\operatorname{End}_R(M)$ and a generating set $\{m_1,\dots, m_n\}$ of $M$ we may produce a matrix $A_\phi\in M_{n\times n}(R)$ such that
$$\require{AMScd}
\begin{CD}
R^n @>A_\phi>> R^n \\
@V\pi VV @VV\pi V\\
M @>>\phi > M
\end{CD}
$$
commutes, it is not the case that in general a matrix $B\in M_{n\times n}(R)$ induces a well-defined endomorphism of $M.$ However, this doesn't mean we can't use the main idea that $(xI - A)^{\textrm{adj}}(xI -A) = \det(xI - A)I\in M_{n\times n}(R[x]),$ we just need to be careful.
First, let's choose our generating set $\{m_1,\dots, m_n\}$ of $M$ and our matrix representation $A_\phi$ of $\phi$ with respect to this generating set. Explicitly, we have some collection of constants $r_{ij}\in R$ such that
$$
\phi(m_i) = \sum_{j=1}^n r_{ij} m_j.
$$
If we let $\delta_{ij} = \begin{cases} 1,\quad i = j\\ 0,\quad i\neq j\end{cases}$ and we consider $M$ as an $R[x]$-module where $x$ acts on $M$ by $xm = \phi(m),$ then the previous equation is equivalent to
$$
\sum_{j}(x\delta_{ij} - a_{ij})m_j = 0.
$$
Observe that if we assemble the coefficients of the $m_j$ as we range over all $j$ and all $i$ into a matrix, we obtain
$$(x\delta_{ij} - a_{ij})_{ij} = xI - A_\phi.$$
Now we apply the adjugate trick. Write $(xI - A_\phi)^{\textrm{adj}} = (b_{ij})_{ij}.$ Then the fact that $(xI - A_\phi)^{\textrm{adj}}(xI - A_\phi) = \det(xI - A_\phi) I$ means that
$$
\sum_{k=1}^n b_{ik}(x\delta_{kj} - a_{kj}) = \det(xI - A_\phi)\delta_{ij}.
$$
Taking our equation $0 = \sum_{j}(x\delta_{kj} - a_{kj})m_j$ and multiplying by $b_{ik},$ we have
$$
0 = \sum_j b_{ik}(x\delta_{kj} - a_{kj})m_j.
$$
Next we sum these equations over $k$:
\begin{align*}
0 &= \sum_{k=1}^n\sum_{j=1}^n b_{ik}(x\delta_{kj} - a_{kj})m_j\\
&=\sum_{j=1}^n\sum_{k=1}^n b_{ik}(x\delta_{kj} - a_{kj})m_j\\
&= \sum_{j=1}^n\det(xI - A_\phi)\delta_{ij} m_j\\
&= \det(xI - A_\phi)m_i.
\end{align*}
This holds for any $i,$ so that $\det(xI - A_\phi) = p(x)$ acts on $M$ identically as zero; i.e., $p(\phi) : M\to M$ is zero.
Best Answer
Let $v$ be a nonzero vector in $V=k^n$. Let $m$ be the least positive integer with $A^mv$ linearly dependent on $v$, $Av,\ldots, A^{m-1}v$. Then there is a monic $f$ of degree $m$ with $f(A)v=0$. Therefore $f(A)A^jv=A^jf(A)v=0$ for $0\le j\le m - 1$ so that $f(A)$ has nullity $\ge m$. Let $W=f(A)V$. The dimension of $W$ is $\le n-m$. Then $A$ acts on $W$ and inductively $g(A)W=0$ for some monic polynomial $g$ of degree $\le n-m$. Therefore $gf$ has degree $\le n$ and annihilates $V$.