What are we solving with row reduction? Why are row reductions so useful in linear algebra? It is easy to get lost in the mechanical solving of equations. I know I can get a matrix into reduced row echelon form. But what are the outcomes of this? What can this mean?

The main point of row operations is that they do not change the solution set of the underlying linear system.  So when you take a system of linear equations, write down its (augmented) coefficient matrix, and row reduce that matrix, you get a new system of equations that has the same solutions as the original system.
When the coefficient matrix is in reduced row echelon form, it is particularly easy to understand the solution set of the linear system.  In particular, you can readily tell whether the system is consistent or inconsistent, and when the system is consistent, you can readily see how a choice for the values of the free (non-pivotal) variables leads to a solution.
I recommend David Lay's book as a textbook for this material.
One thing that sometimes gets lost in the process of Gaussian elimination (the usual algorithm for putting a matrix into reduced row echelon form) is just how similar this process is to the approach, often taught in high school algebra, of solving for one variable in a particular equation and then eliminating its occurrences from all the other equations.  When you produce a pivot in a row of the coefficient matrix, this corresponds to solving the equation for that variable.  When you add or subtract (multiples of) this pivotal row from the other rows to produce zeros in the pivotal column, the new rows represent the equations you get by substituting in for the pivotal variable.  In a sense, Gaussian elimination is just an excellent way to keep track of the process of solving for variables and then eliminating them from the other equations.
Here's an example.  We start with the equations
$3x + 6y + 3z = 3$
$2x + y  + 7z  = 6$.
The corresponding matrix is
$\left[\begin{array}{rrr|r} 3 & 6 & 3 & 3\\\
 2 & 1 & 7 & 6\end{array}\right]$.
Dividing the first row/equation by 3 gives you the equation $x + 2y + z = 1$, which you can think of as $x = 1 - 2y - z$.  You can plug that expression into the second equation to get
$2(1-2y -z) + y + 7z = 6$, which simpifies to $-3y + 5z = 4$.  Notice that when you row reduce by dividing the first row by 3 and then subtracting twice row 1 from row 2, you get the same thing:
$\left[\begin{array}{rrr|r} 3 & 6 & 3 & 3\\\
 2 & 1 & 7 & 6\end{array}\right] \to  \left[\begin{array}{rrr|r} 1 & 2 & 1 & 1\\\
 2 & 1 & 7 & 6 \end{array}\right] \to  \left[\begin{array}{rrr|r} 1 & 2 & 1 & 1\\\
 0 & -3 & 5 & 4\end{array}\right].$
