How to determine sign of coefficients in simple spring, damper, mass system? For a system of the sort shown below: 

I have come to realize that I continuously make mistakes when it comes to determining the signs (or specifically the direction of the forces) of the coefficients in the governing differential equation. 
What are the steps for solving a problem of this sort while maintaining consciousness of the signs needed to resolve the correct solution? 
I normally go about something like this by looking at each mass in the system and the associated forces acting on it. The forces associated with anchored components are trivial, but I am always making mistakes when it comes to the less obvious ones, like the force from spring kd. 
For instance, the solution to this problem in an unreduced form is supposedly:
$$
m_c \cdot x_p'' + b_s \cdot x_p' = b_d \cdot (x_{in} - x_{p})' + k_d \cdot (x_{in} - x_p)
$$
Where ' represents derivative order. 


*

*Firstly, what is the equal sign symbolic of? Why do we need to relate the forces on the mass to the other forces, and why through an equivalency? Is this assumed to be a static system? 

*Why do we take $x_{in} - x_p$, rather than the other way round in each circumstance? 

*Why do we not have a negative coefficient (in this case) in front of the $k_d$ spring constant while at other times we may?    
 A: Always set up a coordinate system, defining positive directions. Let's choose + to the right for both $x_{in}$ and $x_p$. You didn't give it, but let's also assume that the spring is relaxed when $x_{in}=x_{p}$.

Firstly, what is the equal sign symbolic of?

Next, always write the sum of the force terms on one side and $m\ddot{x}$ (I'll use dot notation for time derivatives) on the other.  That helps keep the negative sign difficulties to a minimum.  $$\sum_i F_i = m\ddot{x}$$. Later you can rearrange terms if you want to.  So that explains the equal sign.

Why do we need to relate the forces on the mass to the other forces, and why through an equivalency?

The left side of my equation sums all the forces acting on the mass: spring, damping, and friction.  The only ``equivalency'' is that the sum of the forces must be isomorphic with $m\ddot{x}$.

Is this assumed to be a static system? 

No, it's looks like its assumed to be moving because of the sliding friction term. If you assume a static system, the acceleration and velocities will be zero. BTW, that seems to have a funny form to me because the solution you state has made the sliding friction force proportional to the velocity. Most 1st order sliding friction terms are constant in magnitude.
To answer your other questions, let's consider the sum of the forces term:
The spring force: If $x_{in}-x_{p} >0$, the spring pushes the mass to the right which is a positive directed force, so we add $k_d(x_{in}-x_{p})$ in the sum.
The damping force This will be proportional to the velocity of the damper. If $\dot{x}_{in}>0$, this results in the damper pushing on the mass $m$ to the right. If $\dot{x}_p>0$, this results in the damper pulling on the mass $m$ to the left. So we add $b_d\dot{x}_{in}$ and subtract $b_d\dot{x}_{p}$ in the sum.
The sliding friction force Again, this is weird to me, but it appears the magnitude is going to be $b_s\dot{x}_p$.  Now, why not $\dot{x}_{in}-\dot{x}_p$? Because, unlike the damper and the spring, the motion of $x_{in}$ isn't directly affecting the friction. Now, what about the sign: the sign of the friction term will be opposite the sliding direction, which is given by the sign of $\dot{x}_p$, so we subtract $b_s\dot{x}_p$.
Now you can rearrange your terms and get the given solution. But always start these problems with Newton's 2nd Law.
