Let $(R, \mathfrak m, k)$ be a regular local ring of dimension $2$. Note that Hom$_R (k, M)$ is a finite dimensional $k$-vector space for any finitely generated $R$-module $M$ and Hom$_R (k, R/I) \cong (I:\mathfrak m)/I$ for any ideal $I$.
Then, is it true that $1+ \dim_k$ Hom$_R(k, R/I)=\mu (I) $ ?
Note that $\mu (I)=\dim_k I/I\mathfrak m$.
Also, note that if $\mu(I)=n$, then $I$ has a minimal free resolution $0\to R^{n-1}\to R^n \to I\to 0$ . So in particular one also has a minimal free resolution $0\to R \to R^2 \to \mathfrak m \to 0$ . I'm not sure whether these are helpful in the context or not.
This is correct. Starting with $0\to I\to R\to R/I\to 0$, you get $\mathrm{Hom}(k,R/I)=\mathrm{Ext}^1(k,I)$, since $\mathrm{Ext}^1(k,R)=0$. From the exact sequence you have written, one has $0\to\mathrm{Ext}^1(k,I)\to\mathrm{Ext}^2(k,R^{n-1})\to\mathrm{Ext}^2(k,R^n)$. The last map is zero, since all are $k$-vector spaces and the minimality of the resolution says that all the entries in the map (thought of as a matrix over $R$, say) are in the maximal ideal. Now, $\mathrm{Ext}^2(k,R)=k$ and so you get $\dim_k\mathrm{Hom}(k,R/I)=\dim_k\mathrm{Ext}^1(k,I)=\dim_k\mathrm{Ext}^2(k,R^{n-1})=n-1$.