From your notation I assume that your true model is:
$ Y_i=\beta_1+\beta_2 X_i + \epsilon_i \qquad i=1,\ldots,n $ where $\beta_1$ and $\beta_2$ are the true parameters of the model. Then, if you assume that the (true) error term is iid and has zero conditional expectation you get (i.e., $E[\epsilon_i|X]=0$):
$ E[Y_i|X_i]=\beta_1 + \beta_2 X_i + E[\epsilon_i|X_i]\\ =\beta_1+\beta_2 X_i $ which is what you want to estimate using OLS. Let $b_1$ and $b_2$ be the OLS estimates of $\beta_1$ and $\beta_2$ respectively. Then, $ b_2=\frac{\sum_{i=1}^{n} (Y_i-\bar{Y})(X_i-\bar{X})}{\sum_{i=1}^{n}(X_i-\bar{X})^2}\\ $ Since, $ \sum_{i=1}^{n} (Y_i-\bar{Y})(X_i-\bar{X})=\sum_{i=1}^{n} Y_i(X_i-\bar{X})-\bar{Y}\sum_{i=1}^{n}(X_i-\bar{X})\\ =\sum_{i=1}^{n} Y_i(X_i-\bar{X}) $ and, $ \sum_{i=1}^{n} (X_i-\bar{X})(X_i-\bar{X})=\sum_{i=1}^{n} X_i(X_i-\bar{X})-\bar{X}\sum_{i=1}^{n}(X_i-\bar{X})\\ =\sum_{i=1}^{n} X_i(X_i-\bar{X}) $ because $\sum_i(X_i-\bar{X})=0$ you can rewrite $b_2$ as follows: $ b_2=\frac{\sum_{i=1}^{n} Y_i(X_i-\bar{X})}{\sum_{i=1}^{n}X_i(X_i-\bar{X})}\\ $ Let $e_i$ be the OLS residual, i.e., $e_i=Y_i-b_1-b_2X_i$, $i=1,\ldots,n$. Then, $ \sum_i(X_i-\bar{X})e_i=\sum_i(X_i-\bar{X})(Y_i-b_1-b_2X_i)\\ =\sum_i Y_i(X_i-\bar{X})-b_1\sum_i(X_i-\bar{X})-b_2\sum_i X_i(X_i-\bar{X})\\ =\sum_i Y_i(X_i-\bar{X})-b_2\sum_i X_i(X_i-\bar{X})\\ =\sum_i Y_i(X_i-\bar{X})-\left(\frac{\sum_i Y_i(X_i-\bar{X})}{\sum_i X_i(X_i-\bar{X})}\right)\sum_i X_i(X_i-\bar{X}) $ where again I have used the fact that $\sum_i (X_i-\bar{X})=0$.