There are many different ways to look at degrees of freedom. I wanted to provide a rigorous answer that starts from a concrete definition of degrees of freedom for a statistical estimator as this may be useful/satisfying to some readers:
Definition: Given an observational model of the form $$y_i=r(x_i)+\xi_i,\ \ \ i=1,\dots,n$$ where $\xi_i=\mathcal{N}(0,\sigma^2)$ are i.i.d. noise terms and the $x_i$ are fixed. The degrees of freedom (DOF) of the estimator $\hat{y}$ is defined as $$\text{df}(\hat{y})=\frac{1}{\sigma^2}\sum_{i=1}^n\text{Cov}(\hat{y}_i,y_i)=\frac{1}{\sigma^2}\text{Tr}(\text{Cov}(\hat{y},y)),$$ or equivalently by Stein's lemma $$\text{df}(\hat{y})=\mathbb{E}(\text{div} \hat{y}).$$
Using this definition, let's analyze linear regression.
Linear Regression: Consider the model $$y_i=x_i\beta +\xi_i,$$ with $x_i\in\mathbb{R}^p$ are independent row vectors. In your case, $p=2$, and the $x_i={z_i,1}$ correspond to a point and the constant $1$, and $\beta=\left[\begin{array}{c}
m\\
b
\end{array}\right]$, that is a slope and constant term so that $x_i \beta=m z_i+b$. Then this can be rewritten as $$y=X\beta+\xi$$ where $X$ is an $n\times p$ matrix whose $i^{th}$ row is $x_i$. The least squares estimator is $\hat{\beta}^{LS}=(X^T X)^{-1}X^Ty$. Let's now based on the above definition calculate the degrees of freedom of $SST$, $SSR$, and $SSE$.
$SST:$ For this, we need to calculate $$\text{df}(y_i-\overline{y})=\frac{1}{\sigma^2}\sum_{i=1}^n\text{Cov}(y_i-\overline{y},y_i)=n-\frac{1}{\sigma^2}\sum_{i=1}^n\text{Cov}(\overline{y},y_i)=n-\frac{1}{\sigma^2}\sum_{i=1}^n \frac{\sigma^2}{n}=n-1.$$
$SSR:$ For this, we need to calculate $$\text{df}(X\hat{\beta}^{LS}-\overline{y})=\frac{1}{\sigma^2}\text{Tr}\left(\text{Cov}(X(X^TX)^{-1}X^y,y\right)-\text{df}(\overline{y})$$ $$=-1+\text{Tr}(X(X^TX)^{-1}X\text{Cov(y,y)})$$ $$=-1+\text{Tr}(X(X^TX)^{-1}X^T)$$ $$=p-1.$$ In your case $p=2$ since you will want $X$ to include the all ones vector so that there is an intercept term, and so the degrees of freedom will be $1$. However note that this will equal the number of parameters when we are doing regression with multiple parameters.
$SSE:$ $(n-1)-(p-1)=n-p$, which follows linearity of $df$.
The principle underlying least squares regression is that the sum of the squares of the errors is minimized. We can use calculus to find equations for the parameters $\beta_0$ and $\beta_1$ that minimize the sum of the squared errors.
Let $S = \displaystyle\sum\limits_{i=1}^n \left(e_i \right)^2= \sum \left(y_i - \hat{y_i} \right)^2= \sum \left(y_i - \beta_0 - \beta_1x_i\right)^2$
We want to find $\beta_0$ and $\beta_1$ that minimize the sum, $S$. We start by taking the partial derivative of $S$ with respect to $\beta_0$ and setting it to zero.
$$\frac{\partial{S}}{\partial{\beta_0}} = \sum 2\left(y_i - \beta_0 - \beta_1x_i\right)^1 (-1) = 0$$
notice that this says,
$$\begin{align}\sum \left(y_i - \beta_0 - \beta_1x_i\right) &= 0 \\
\sum \left(y_i - \hat{y_i} \right) &= 0 \qquad (eqn. 1)\end{align}$$
Hence, the sum of the residuals is zero (as expected). Rearranging and solving for $\beta_0$ we arrive at,
$$\sum \beta_0 = \sum y_i -\beta_1 \sum x_i $$
$$n\beta_0 = \sum y_i -\beta_1 \sum x_i $$
$$\beta_0 = \frac{1}{n}\sum y_i -\beta_1 \frac{1}{n}\sum x_i $$
now taking the partial of $S$ with respect to $\beta_1$ and setting it to zero we have,
$$\frac{\partial{S}}{\partial{\beta_1}} = \sum 2\left(y_i - \beta_0 - \beta_1x_i\right)^1 (-x_i) = 0$$
and dividing through by -2 and rearranging we have,
$$\sum x_i \left(y_i - \beta_0 - \beta_1x_i\right) = 0$$
$$\sum x_i \left(y_i - \hat{y_i} \right) = 0$$
but, again we know that $\hat{y_i} = \beta_0 + \beta_1x_i$. Thus, $x_i = \frac{1}{\beta_1}\left( \hat{y_i} - \beta_0 \right) = \frac{1}{\beta_1}\hat{y_i} -\frac{\beta_0}{\beta_1}$. Substituting this into the equation above gives the desired result.
$$\sum x_i \left(y_i - \hat{y_i} \right) = 0 $$
$$\sum \left(\frac{1}{\beta_1}\hat{y_i} - \frac{\beta_0}{\beta_1}\right) \left(y_i - \hat{y_i} \right) = 0$$
$$\frac{1}{\beta_1}\sum \hat{y_i} \left(y_i - \hat{y_i} \right) - \frac{\beta_0}{\beta_1} \sum \left(y_i - \hat{y_i} \right)= 0$$
Now, the second term is zero (by eqn. 1) and so, we arrive immediately at the desired result:
$$\sum \hat{y_i} \left(y_i - \hat{y_i} \right) = 0 \qquad (eqn. 2)$$
Now, let's use eqn. 1 and eqn. 2 to show that
$\sum \left(\hat{y_i} - \bar{y_i} \right) \left( y_i - \hat{y_i} \right) = 0$ - which was your original question.
$$\sum \left(\hat{y_i} - \bar{y_i} \right) \left( y_i - \hat{y_i} \right) = \sum \hat{y_i} \left( y_i - \hat{y_i} \right) - \bar{y_i} \sum \left( y_i - \hat{y_i} \right) = 0$$
Best Answer
You are close. I presume that from $SSE = SST - SSR$ you have $$\sum_i e_i^2 = \sum_i (y_i - \bar{y})^2 - \sum_i (\hat{y}_i - \bar{y})^2 = \sum_i \hat{y}_i^2 - \sum_i y_i^2 - 2 \sum_i (y_i - \hat{y}_i) \bar{y},$$ and are wondering what to do with the extra term.
It turns out the last term is zero; you can prove this by looking at the "normal equations" (i.e. look at the derivation of the least squares coefficients) or by laboriously plugging in the definition of $\hat{y}_i$.
Note that the result holds even without the expectations.
[By the way, why do you have an index $i$ in $\bar{y}_i$? Isn't $\bar{y} := \frac{1}{n} \sum_i y_i$?]