Spring damping in Energy Function

The damping force you mentioned $f=-k \frac{\dot{l} \cdot l}{|l|} \frac{l}{|l|}$ is a special case of $f=-k \dot{C} \frac{\partial C}{\partial \mathbf{x}}$.

Let $$ \begin{align} C(\mathbf{x}) &= \lVert\mathbf{l}\rVert-l_0 \\ \mathbf{l} &= \mathbf{x}_1 - \mathbf{x}_2 \\ \mathbf{\dot{l}} &= \mathbf{v}_1 - \mathbf{v}_2 \\ \mathbf{v} &= \dot{\mathbf{x}} = \begin{pmatrix}\mathbf{v}_1 \\ \mathbf{v}_2\end{pmatrix} \end{align} $$ where $l_0$ is the rest length of the spring, and $\mathbf{x}_1$, $\mathbf{x}_2$ are the positions of two end points. The damping force is $$ \begin{align} \mathbf{f} &= -k \dot{C} \frac{\partial C}{\partial \mathbf{x}} \\ &= -k (\frac{\partial C}{\partial \mathbf{x}})^T \dot{\mathbf{x}} \frac{\partial C}{\partial \mathbf{x}}\\ &= -k (\mathbf{l}^T\mathbf{l})^{-1/2} \mathbf{l}^T \begin{pmatrix}\mathbf{I}& -\mathbf{I} \end{pmatrix} \dot{\mathbf{x}} (\mathbf{l}^T\mathbf{l})^{-1/2} \begin{pmatrix}\mathbf{I}& -\mathbf{I} \end{pmatrix}^T \mathbf{l} \\ &= -k (\mathbf{l}^T\mathbf{l})^{-1} \begin{pmatrix}\mathbf{l}^T& -\mathbf{l}^T \end{pmatrix} \mathbf{v} \begin{pmatrix}\mathbf{l} \\ -\mathbf{l} \end{pmatrix}\\ &= -k \frac{(\mathbf{l}\cdot \mathbf{\dot{l}}) }{\lVert \mathbf{l} \rVert^2} \begin{pmatrix}\mathbf{l} \\ -\mathbf{l} \end{pmatrix} \end{align} $$

This matches your damping force equation. The $-\mathbf{l}$ part is just the force for the other particle, which has the opposite direction as expected. Bold letters are used to denote vector quantities.

As mentioned in [baraff98], this damping force is along the direction $\frac{\partial C}{\partial \mathbf{x}}$, and its magnitude is the projected component of velocity $\frac{\partial C}{\partial \mathbf{x}}^T \mathbf{v}=\frac{\partial C}{\partial \mathbf{x}}^T \mathbf{\dot{x}}=\dot{C}$.