The model sum of squares, also known as the explained sum of squares, is the sum of squared deviations of the model's predicted value $\hat{y}_i$ from the outcome variable's unconditional mean $\bar{y}$:
$$ \mathit{ESS} = \sum_{i=1}^n (\hat{y}_i - \bar{y})^2 $$
In linear regression, the total sum of squares equals the explained sum of squares plus the residual sum of squares because the residuals are statistically orthogonal (by construction) to the explanatory variables.
$$ \underbrace{\sum_{i=1}^n (y_i - \bar{y})^2}_{\mathit{TSS}} = \underbrace{\sum_{i=1}^n (\hat{y}_i - \bar{y})^2}_{\mathit{ESS}} + \underbrace{\sum_{i=1}^n (y_i - \hat{y}_i)^2}_{\mathit{RSS}}$$
The residuals' degree of freedom is an entirely different concept. The residuals degree of freedom is the dimension of the linear subspace in which the residual vector lies.
Some intuition and motivation for degrees of freedom
Imagine have some vector $ \boldsymbol{\epsilon} = (x,y,z) \in \mathbb{R}^3$, that is, $\boldsymbol{\epsilon}$ is some point in three dimensional space.
Scenario 1:
Q: You are told $x + y + z = 1$. What's the space of points $(x,y,z)$ that satisfy that constraint?
A: a plane: 
. A plane is a two-dimensional linear subspace. That restriction implies $\boldsymbol{\epsilon}$ lies in a 2 dimensional linear subspace of $\mathbb{R}^3$.
Scenario 2:
Q: You are told $x + y + z = 1$ and that $y + z = 0$. What's the space of points $(x,y, z)$ that satisfy those constraints?
A: It's a line. 
. A line is a one dimensional linear subspace. The restrictions imply $\boldsymbol{\epsilon}$ lies in a one dimensional linear subspace of $\mathbb{R}^3$.
So what's residuals' degrees of freedom?
In linear regression, your residual vector is an $n$ dimensional vector where $n$ is the number of observations.
$$ \boldsymbol{\epsilon} = \begin{bmatrix} \epsilon_1 \\ \epsilon_2 \\ \ldots \\ \epsilon_n \end{bmatrix} $$
So the residual vector $\boldsymbol{\epsilon}$ could take any value in $\mathbb{R}^n$? No!
For each coefficient you estimate, you impose the constraint that the residual vector is orthogonal to associated right hand side variable. If you're running the regression:
$$y_i = b_0 + b_1 x_{i,1} + b_2 x_{i,2} + \ldots + b_k x_{i,k} + $$
you have $k+1$ linear constraints. In matrix form, the $k+1$ equations can be written as $X'X \mathbf{b} = X'\mathbf{y}$. Hence the residual vector is restricted to an $n-k-1$ dimensional linear subspace of $\mathbb{R}^n$.
