I think the relationship might be clearer with a different expression of the model. Consider the logistic regression model as its expressed on p. 225 of this excerpt:
Formally, the model logistic regression model is that
$$\log\frac{p(x)}{1-p(x)} = \beta_0 + \beta_1x$$
Equivalently:
$$\frac{p(x)}{1-p(x)} = e^{\beta_0 + \beta_1x}$$
In words, logistic regression models log odds as a linear function of the predictors; the odds are the exponentiation of this linear combination. In the multivariate case, this gives:
$$\frac{p(x)}{1-p(x)} = e^{\beta_0 + \beta_1x_1...+\beta_nx_n}$$
Look at it as a product
$$\frac{p(x)}{1-p(x)} = e^{\beta_0}e^{\beta_1x_1}...e^{\beta_nx_n}$$
and the relationship between the fitted coefficients and predictors is clear: For every one-unit increase in $x_i$, the log-odds increase by a factor of $e^{\beta_i}$. This seems to be the relationship that your formula seeks to stress. I believe the proof you ask for is this:
$$OR = \frac{\operatorname{odds}(F(\hat{\textbf x}))}{\operatorname{odds}(F(\textbf x))} = \frac{\frac{F(\hat{\textbf x})}{1 - F(\hat{\textbf x})}}{\frac{F(\textbf x)}{1 - F(\textbf x)}} = \frac{e^{\beta_0 + \beta_1 x_1 + ... + \beta_m(x_m+1)+\beta_nx_n}}{e^{\beta_0 + \beta_1 x_1 + ... + \beta_mx_m+\beta_nx_n}} = e^{\beta_m}$$
with $\hat{x}_i = x_i$ for $i \ne m$, and $\hat x_m = x_m + 1$. (For my part, I don't find this particularly intuitive or illustrative.)
What probabilities are being compared?
If $x_m$ is continuous, it's comparing the odds with a one-unit change in $x_m$, while holding all other predictors equal; if $x_m$ is binary, it's comparing two cases in which all variables are equal except the presence or absence of $x_m$.
Is it really still the OR once it changes?
You may find it more helpful to think of it as your understanding of the odds ratio changing in light of new information.
