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 $df$.
A first answer is that usually, the units of one variable are not the units in the other variabl. For example when you have pairs $(size_k, weight_k), k=1\cdots n$, among a certain human population sample represented as points, depending on the units you take on each axis, orthogonal projection of the points on the best fit line for a certain choice of units will not fall at the same place if you change the units say on one of the axes.
Best Answer
You need a bit of undergraduate calculus to understand this.
Let $y=mx+b$ denote the equation of the line which minimizes
$$ S=\sum_{k=1}^n[y_k-(mx_k+b)]^2$$
Then S is a second-degree polynomial in two variables $m$ and $b$. So the partials derivatives of $S$ with respect to both $m$ and $b$ will be zero at any extreme value, such as the minimum.
Taking $\dfrac{\partial S}{\partial b}$ we get
\begin{eqnarray} \frac{\partial S}{\partial b}&=&-2\sum_{k=1}^n[y_k-(mx_k+b)]\\ &=&0 \end{eqnarray}
Therefore,
\begin{eqnarray} \sum_{k=1}^ny_k&=&m\sum_{k=1}^nx_k+\sum_{k=1}^nb\\ \sum_{k=1}^ny_k&=&m\sum_{k=1}^nx_k+nb\\ \frac{1}{n}\sum_{k=1}^ny_k&=&\frac{m}{n}\sum_{k=1}^nx_k+b\\ \bar{y}&=&m\bar{x}+b \end{eqnarray}