Would the following be the kind of counterexample you have in mind?
Let $k=\mathbf{Z}/2\mathbf{Z}$ be the field of two elements, and let $m=3$. Let further $V$ be a free $R$-module of rank $n=1$, so as a vector space over $k$ we have $V\cong k^3$. Let $W$ be the zero sum $k$-subspace of $V$: $W=\{(a_1,a_2,a_3)\in k^3\mid a_1+a_2+a_3=0\}=\{000,110,101,011\}.$ We easily see that $W$ generates $V$ as an $R$-module, because for each coordinate position there is a vector in $W$ such that its component in that position is non-zero. Yet no element of $W$ generates $V$ alone as an $R$-module. Any element has at least a single zero component, so the cyclic $R$-module generated by that element cannot be all of $V$.
Edit1: If $k=\{x_1,x_2,\ldots,x_q\}$ then the two-dimensional subspace $W\subseteq k^{q+1}$ spanned by the vectors $\vec{a}=(0,1,1,1,1,\ldots,1)$ and $\vec{b}=(1,x_1,x_2,\ldots,x_q)$ has the property that any vector of $W$ has at least a single component equal to zero. This is because a non-zero scalar multiple of $\vec{b}$ has all the $q$ elements permuted in the last $q$ positions, so one of them will get cancelled, when we add a non-zero multiple of $\vec{a}$. Also, obviously all the positions have something non-zero in either $\vec{a}$ or $\vec{b}$. The argument works the same as in the earlier case $q=2$.
I don't know yet how to prove that $m=q+1$ is the shortest length, where such a subspace $W$ can be found. I think that is the case, though. Two remarks:
1) We can make vectors of $W$ longer by replicating one of the coordinates as many times as we need.
2) Does this show that counterexamples with a 2-dimensional $k$-space $W$ exist, whenever $mn\ge |k|+1$? This is wrong, but in the case $n=1$ we do get counterexamples like this, if $m\ge |k|+1$.
Edit2:
Ok, here's the missing part.
Lemma. Assume $|k|=q$, and $W$ is an $\ell$-dimensional subspace of $k^m$ such that for all the $m$ coordinate positions there is a vector $w\in W$ with a non-zero component in that position, but also every vector of $W$ has at least one coordinate equal to zero. Then $\ell\ge2$ and $m\ge q+1$.
Proof. Obviously $\ell=1$ doesn't work, so $\ell\ge2$. Let $S_i$ be the subspace of $W$ consisting of those vectors that have a zero in position $i$. Clearly $\dim_k S_i=\ell-1$, so $|S_i|=q^{\ell-1}$. From our assumptions it follows that the union of all the subsets $S_i$ covers all of $W$. OTOH the zero vector is contained in all of the sets $S_i$, so there is some overlap. Therefore $ \sum_i|S_i|= mq^{\ell-1}>|W|=q^\ell. $ For this to hold we must have $m\ge q+1.$ Q.E.D.
2') Doesn't it follow that counterexamples exist, iff $m\ge|k|+1$?