In the context of a Bernoulli distribution, what exactly the relationship is between the linear predictor and the mean of the distribution function? A Bernoulli distribution can be expressed as
$$
{\displaystyle f(k;p)=p^{k}(1-p)^{1-k}\quad {\text{for }}k\in \{0,1\}}
$$
its mean is $p$. Let p = 0.7, the Bernoulli distribution can be expressed as
$$
{\displaystyle f(k;0.7)=0.7^{k}(1-0.7)^{1-k}\quad {\text{for }}k\in \{0,1\}} \tag{1}
$$ 
per wiki

The link function provides the relationship between the linear
  predictor and the mean of the distribution function.

for the Bernoulli distribution, the mean is p = 0.7
the logit function is in this form
$${\displaystyle \operatorname {logit} (p)=\log \left({\frac {p}{1-p}}\right)}$$
this equation
$${\displaystyle \mathbf {X} {\boldsymbol {\beta }}=\ln \left({\frac {\mu }{1-\mu }}\right)\,\!}$$
comes from Common distributions with typical uses and canonical link functions table on wiki.
for the case of Bernoulli distribution with p = 0.7
$$\mathbf {X} {\boldsymbol {\beta }} = \log \left({\dfrac {p}{1-p}}\right)$$
The Mean function 
$$
{\displaystyle \mu ={\frac {\exp(\mathbf {X} {\boldsymbol {\beta }})}{1+\exp(\mathbf {X} {\boldsymbol {\beta }})}}={\frac {1}{1+\exp(-\mathbf {X} {\boldsymbol {\beta }})}}\,\!} \tag{2}
$$
also comes from Common distributions with typical uses and canonical link functions table on wiki.
for the case of Bernoulli distribution with p = 0.7
$$
{\displaystyle p ={\frac {\exp(\mathbf {X} {\boldsymbol {\beta }})}{1+\exp(\mathbf {X} {\boldsymbol {\beta }})}}={\frac {1}{1+\exp(-\mathbf {X} {\boldsymbol {\beta }})}}\,\!} \tag{3}
$$
does this mean that Equation 2 would output 0.7 for a a Bernoulli distribution parameterized by Equation 1? If yes, what is the detailed procedure? 
 A: First to clear things out:
If the response is $Y \sim \text{Normal}(\mu, \sigma^{2})$ we can model how the mean $\text{E}(Y) = \mu$ depends on some variables with equation:
$$
\mu = \mathbf{x}^{T} \cdot \mathbf{\beta}
$$
and because right hand side is a linear function of $\mathbf{\beta}$:
$$
\mu = \beta_{0} + \beta_{1} \cdot x_{1} + ... + \beta_{m} \cdot x_{m}
$$
we call this linear regression. 
Now if the response is $Y \sim \text{Bernoulli}(\pi)$ we can model how the mean $\text{E}(Y) = \pi$ depends on some variables with equation:
$$
\pi = \frac{\exp(\mathbf{x}^{T} \cdot \mathbf{\beta})}{1 + \exp(\mathbf{x}^{T} \cdot \mathbf{\beta})}
$$
and because the right hand side is called (standard) logistic function or sigmoid:
$$
S(x) = \frac{\exp(x)}{1 + \exp(x)}
$$
we call this logistic regression. The inverse of $S(x)$ is a function
$$
\text{logit}(x) = \ln\left( \frac{x}{1 - x} \right)
$$
called logit because when $x = \text{Pr}(A)$ is probability of some event A this function returns logarithm of odds for the event A. 
Now in this case, when the response is Bernoulli random variable, logit links the mean to linear predictor $\mathbf{x}^{T} \cdot \mathbf{\beta}$:
$$
\text{logit}(\pi) = \text{logit}(S(\mathbf{x}^{T} \cdot \mathbf{\beta})) = \mathbf{x}^{T} \cdot \mathbf{\beta}
$$
and we say that logit is the link function for Bernoulli response. 
Now to answer your question:
Let's say we have only one explanatory random variable $X$ and we would like to know how mean $\pi$ depends on $X$:
$$
\pi(X) = \text{Pr}(Y = 1 | X)
$$
Using logistic regression:
$$
\pi(X) = S(\mathbf{x}^{T} \cdot \mathbf{\beta})
$$
the function $S(\mathbf{x}^{T} \cdot \mathbf{\beta})$ will return an estimate of $\pi$ given value $x$ of $X$ and given an estimate of $\mathbf{\beta}$. 
To give you detailed procedure and to simplify let's say we have no explanatory variables ($m = 0$), so $\mathbf{x}^{T} \cdot \mathbf{\beta}$ is just $\beta_{0}$ and model becomes:
$$
 \pi = S(\beta_{0}) = \frac{\exp(\beta_{0})}{1 + \exp(\beta_{0})}
$$
To estimate $\beta_{0}$ we can use maximum likelihood (MLE) method and using Bernoulli probability mass function for sample size $n$ we get the likelihood in terms of $\pi$:
$$
\mathcal{L}(\pi | \mathbf{y}) = \prod_{i = 1}^{n} \pi^{y_{i}} (1 - \pi)^{1 - y_{i}} \qquad \text{where } y_{i} \sim \text{iid Bernoulli}(\pi)
$$
and taking into account our model we get the likelihood in terms of $\beta_{0}$:
$$
\mathcal{L}(\beta_{0} | \mathbf{y}) = \prod_{i = 1}^{n} \left(\frac{\exp(\beta_{0})}{1 + \exp(\beta_{0})}\right)^{y_{i}} \left(1 - \frac{\exp(\beta_{0})}{1 + \exp(\beta_{0})}\right)^{1 - y_{i}}
$$
using standard procedure of taking the logarithm on both sides to get log-likelihood and then differentiate the log-likelihood with respect to $\beta_{0}$ and equate to zero, we finally get an estimate:
$$
\hat{\beta_{0}} = \ln\left(\frac{\frac{1}{n}\sum_{i=1}^{n} {y_{i}}}{1 - \frac{1}{n}\sum_{i=1}^{n} {y_{i}}}\right)
$$
which is logit of $\frac{1}{n}\sum_{i=1}^{n} {y_{i}}$:
$$
\hat{\beta_{0}} = \text{logit} \left( \frac{1}{n}\sum_{i=1}^{n} {y_{i}} \right)
$$
so we see the logistic function $S(\beta_{0})$ will return 
$$
S(\hat{\beta_{0}}) = \frac{1}{n}\sum_{i=1}^{n} {y_{i}}
$$
which is maximum likelihood estimator of $\pi$.
To answer your comment:
When using this linear model:
$$
\text{logit}(\pi) = \mathbf{x}^{T} \cdot \mathbf{\beta}
$$
we assume that logarithm of odds of $Y = 1$ is equal to $\mathbf{x}^{T} \cdot \mathbf{\beta}$ for some unknown vector of parameters $\mathbf{\beta}$.
Because the standard logistic function $S$ is the inverse of logit function, this formula: 
$$
S(\text{logit}(\pi)) = \pi
$$
holds. In other words: the logistic function $S$ maps logarithm of odds $\text{logit}(\pi)$ to probability $\pi$. 
In practice, this linear model $\mathbf{x}^{T} \cdot \mathbf{\beta}$ is only an approximation of $\text{logit}(\pi)$ and we can only estimate $\mathbf{\beta}$ from sample of values $y_{i}$ of Bernoulli response and corresponding values of explanatory variables $x_{i}$ to get $\mathbf{\hat{\beta}}$. So this formula:
$$
\pi \approx \frac{\exp(\mathbf{x}^{T} \cdot \mathbf{\hat{\beta}})}{1 + \exp(\mathbf{x}^{T} \cdot \mathbf{\hat{\beta}})}
$$
holds only approximately and the right hand side of this formula is called an estimator $\hat{\pi}$ of $\pi$. 
Also if we have at least one explanatory variable $X$, this means that vector $x$ is not equal to a constant 1: $\mathbf{x} \neq 1$, then $\pi$ in this equation represents conditional mean:
$$
\pi = \text{E}(Y | X) = \text{Pr}(Y = 1 | X)
$$ 
and does not represent unconditional mean: $\text{E}(Y)$.
