I found this question somewhat confusing, because of the OP's expectations, and the possible interpretation of the term torsion. For this reason I am giving an answer. It may be more geometric than the OP wants, but my own view is that thinking geometrically makes the makes the answers to these sorts of questions clear, while thinking algbraically can be confusing.
Let's suppose that $R$ (and hence $R_{\mathfrak m}$) are integral domains. A module over an integral domain is called torsion if every element of the module is annihilated by a non-zero element of the domain. If the module is finitely generated, then choosing such a non-zero element of the domain for each generator and multiplying them together, we find a single non-zero element of the domain which annihilates the entire module.
Now if $M$ is a f.g. module over a $A$, and $I$ is its annihilator, then $M_{\mathfrak p}$ is non-zero for a point $\mathfrak p \in $ Spec $A$ if and only if $\mathfrak p \supset I$, and so the support of $M$ is equal to the closd subset $V(I)$ of Spec $A$.
Combining the two paragraphs, we are considering a f.g. module over $R_{\mathfrak m}$ whose support is contained in $V(f)$ for some non-zero $f \in R_{\mathfrak m}$,
and asking if its support as an $R$-module consists of the single point $\mathfrak m$. We may as well make the support in Spec $R_{\mathfrak m}$ as large as possible, to try and make the support
in Spec $R$ as large as possible, and so let's assume the support actaully is $V(f)$ (e.g. by taking $M = R/\mathfrak m$). Clearing denominators, we may also assume that $f \in R$. In fact, we should assume $f \in \mathfrak m$, so that it does not become a unit in $R_{\mathfrak m}$. (Otherwise $M$ would be annihilated by a unit of $R_{\mathfrak m}$, and hence would equal $0$.)
Our question then comes down to asking, for a non-zero $f \in \mathfrak m$, whether or not the image of $V(f) \subset$ Spec $R_{\mathfrak m}$ necessarily
equals the single point $\mathfrak m$. The answer is no in general,
which means that the support of $M$, thought of as an $R$-module, will typically be larger than just $\mathfrak m$.
The reason is that $V(f)$ (in Spec $R_{\mathfrak m}$) is the germ at $\mathfrak m$ of the hypersurface in Spec $R$ cut out by $f$ (i.e. it is the germ at $\mathfrak m$ of the vanishing locus of $f$ in Spec $R$), so its Zariski closure in Spec $R$ will be the irreducible component of that hypersurface passing through $\mathfrak m$. (Let's assume that $R$ is Noetherian, so that irreducible components exist and are well behaved.)
So the only way that the image of $V(f)$ could be just $\mathfrak m$ is if $\mathfrak m$ itself were a hypersurface in Spec $R$. This will be possible if $R$ has dimension one, but only in that case.
So if $R$ is one-dimensional, then the image of $V(f)$ will be just $\mathfrak m$, but not otherwise.
On the other hand, if we assume that $M$ is torsion in a very strong sense, namely that every element is annihilated by a power of $\mathfrak m$, then its support will simply be $\mathfrak m$, and the image of in Spec $R$ will again be $\mathfrak m$. So in this case $M$ will be supported at $\mathfrak m$ as an $R$-module.
The point is that looking at $R_{\mathfrak m}$ is not at all the same as looking at $R/\mathfrak m$, or $R/\mathfrak m^n$. The latter two rings are supported at $\mathfrak m$, but $R_{\mathfrak m}$ sees the whole germ of Spec $R$ near $\mathfrak m$.