Stating your OP generically:
1. Logistic regression without explanatory variables:
$\color{red}{\text{log}} \,\left[\color{blue}{\text{ODDS(p(Y=1))}}\right]=\color{red}{\text{log}}\left(\frac{\hat p\,(Y=1)}{1-\hat p\,(Y=1)}\right) = \hat\beta_o$
$\hat\beta_o$ is the estimated log odds.
It is an intercept only construct. Exponentiating we get
$$\color{blue}{\text{ODDS(Y=1)}} = \frac{p\,(Y=1)}{1\,-\,p\,(Y=1)} = e^{\,\hat\beta_0}$$
$\color{blue}{\large e^{\hat\beta_o}}$ are the $\color{blue}{\text{ODDS}}$.
Translating into probabilities:
$\color{green}{\Pr(Y = 1)} = \frac{\color{blue}{\text{odds(Y=1)}}}{1\,+\,\color{blue}{\text{odds(Y=1)}}}=\frac{e^{\,\hat\beta_0 }}{1 \,+\,e^{\,\hat\beta_0}}$
This is the second calculation in the OP (i.e. the one containing $-1.12546$).
2. Logistic regression with explanatory variable:
$\color{red}{\text{log}} \,\left[\color{blue}{\text{ODDS(p(Y=1))}}\right]=\color{red}{\text{log}}\left(\frac{\hat p\,(Y=1)}{1-\hat p\,(Y=1)}\right) = \hat\beta_o+\hat\beta_1x_1$
$\color{blue}{\text{ODDS(Y=1)}} = \frac{p\,(Y=1)}{1\,-\,p\,(Y=1)} = e^{\,\hat\beta_0+\hat\beta_1x_1} \tag{*}$
Introducing the odds ratio:
If instead of $x_1$ in $(*)$ we have $x_1+1$ - a one-unit increase:
$$\color{blue}{\text{ODDS(Y=1)}} = \frac{p\,(Y=1)}{1\,-\,p\,(Y=1)} = e^{\,\hat\beta_0+\hat\beta_1x_1+ \hat \beta_1}$$
and
$$\color{green}{\text{ODDS RATIO}} = \frac{\color{blue}{\text{odds|}x_1+1}} {\color{blue}{\text{odds|}x_1}}= \frac{e^{\,\hat\beta_0+\hat\beta_1x_1+ \hat \beta_1}}{e^{\,\hat\beta_0+\hat\beta_1x_1}}= e^{\hat\beta_1}$$
$\color{green}{\large e^{\hat\beta_1}}$ is the $\color{green}{\text{ODDS RATIO}}$.
This is the first calculation in the OP.
For every unit increase in $x_1$ the odds increased by $e^{\hat\beta_1}$.
Hence,
$\color{red}{\log}\,[\color{green}{\text{ODDS RATIO}}] = \hat\beta_1$
Best Answer
Personally, I think it is very difficult for people to grasp what does it mean by certain amount of change in odds...
But it is more clear to directly compare the difference of interest: the p's In your case.
At the current value of your predictors, increase 1 unit (can be any change you specify) in the first predictor means increase of
$logit^{-1}(−14.27+3.32(4)+0.88(7)) - logit^{-1}(−14.27+3.32(3)+0.88(7))$
$\approx0.99-0.86 = 0.13$, i.e., it is a 0.13 increase in probability of being 1.
But of course, this varies as the current values of the predictor changes, a common practice (if you are not interested in this specific predictor value) is to let the current predictor value be at the average, which corresponds the biggest change in probability, then you can say, as the predictor values move towards the end, changes in probability are expected to slow down.