No, I am afraid this maximality is not true.
Let us always suppose that $R$ is a domain: else there is no fraction field (although there are substitutes).
First, the result is certainly not true for higher dimensional rings. For example take
$R=\mathbb C[x,y] , m=(x,y) , K=\mathbb C(x,y)$ .
Then you have strict inclusions
$$\mathbb C[x,y]_m \subset \mathbb C[x,y]_m [1/x] \subset \mathbb C (x,y)$$
Hence we see that the result in your case is not purely formal and we must use dimension one. But unfortunately the result may not be true even then, which answers question 1) negatively.
Indeed, just observe in the case $R=\mathbb C[[t^2,t^3]], \; m=(t^2,t^3), \; R=R_m$ the strict inclusions
$$R=\mathbb C[[t^2,t^3]] \subset \mathbb C [[t]]\subset \mathbb C ((t))$$
To end on a cheerful note, your result is indeed true if $R_m$ is a discrete valuation ring.
Here is the proof. Take $a\in K\setminus R_m$ and consider $R_m\subset R_m [a]\subset K$ ( the first inclusion is strict). Since $a$ is not in $R_m$, we know that $1/a$ *is* in $R_m$: this is characteristic of valuation rings. Hence $1/a$ is a fortiori in $R_m[a]$ so that $1\in R[a]$ and we have proved $R[a]=K$.