Interpretation of normal modes from the mathematical formula In the topic of small oscillations, the system below has a normal mode described by:
$$n_{1} = \frac{x1+x2}{2}.$$

This normal mode is represented as the symmetric mode:

In that case, the center of mass moves as a simple harmonic oscillator. However, the picture also shows that both of them start in the same initial conditions and move in phase. My question is where that information is on the normal coordinate $n_{1}$ since I can not relate the normal mode with the picture representing it. Where does it say the blocks must be strechted the same distance in the same direction in the formula of the normal coordinate?
 A: You should notice that the other normal coordinate is implied to be fixed at zero while you consider the motion along the normal coordinate $n_1$.
The normal coordinates of two particles (or blocks in this case) can generally be written as
\begin{align}
n_1 =& a_{11} x_1 + a_{12} x_2, \\
n_2 =& a_{21} x_1 + a_{22} x_2.\label{eq: n1n2}\tag{1}
\end{align}
In your specific case, $a_{11}=1/2$ and $a_{12} =1/2$. I didn't calculate $a_{21}$ and $a_{22}$, but you should be able to do so according to the definition of the normal modes.
The above set of equations can be solved for $x_1$ and $x_2$ in the form of
\begin{align}
x_1 =& b_{11} n_1 + b_{12} n_2, \\
x_2 =& b_{21} n_1 + b_{22} n_2, \label{eq: x1x2}\tag{2}
\end{align}
where $b_{ij}$ are determined by $a_{ij}$. In fact, by writing the above sets of equations by matrices and vectors, you can confirm that
\begin{equation}
\begin{pmatrix}
b_{11} & b_{12} \\
b_{21} & b_{22}
\end{pmatrix}
=
\begin{pmatrix}
a_{11} & a_{12} \\
a_{21} & a_{22}
\end{pmatrix}
^{-1},
\end{equation}
where $A^{-1}$ means the inverse matrix of a matrix $A$.
The first set of equations (\ref{eq: n1n2}) give the coordinate transformation from the coordinates $(x_1,x_2)$ that was convenient for your measurement and other operations to the normal coordinates $(n_1,n_2)$ that is convenient for calculation and some type of interpretation associated with the calculation. The second set (\ref{eq: x1x2}) gives the inverse transformation. That is, once you calculated the time-evolution $n_1(t)$ and $n_2(t)$ individually by solving the differential equations, you can predict the motion $x_1(t)$ and $x_2(t)$ of the respective particles by (\ref{eq: x1x2}).
The differential equations for the normal coordinates are
\begin{equation}
\frac{d^2 n_i}{dt^2} = -\omega_i^2 n_i(t),
\end{equation}
for $i=1,2$, where $\omega_i^2$ are constants, as long as the force acting on the block $l$ is of the form $F_l = - \sum_j c_{lj} x_j$ with some constants $c_{lj}$ as is the case for your problem. The function, $n_2(t) = 0$ for all $t$, is a valid solution for this equation for the initial condition, $n_2(0) =0$ and $[dn_2/dt](0) =0$. Suppose that this condition is satisfied through (\ref{eq: n1n2}) by particular values of $x_j(0)$ and $[dx_j/dt](0)$ ($j=1,2$) which are prepared by your putting hands on the spring-and-mass system at $t=0$. These
$x_j(0)$ and $[dx_j/dt](0)$ ($j=1,2$) also determine the initial values of $n_1(0)$ and $[dn_1/dt](0)$, and hence give a particular solution $n_1(t)$ of the differential equation above. With this $n_1(t)$ and $n_2(t) =0$, through (\ref{eq: x1x2}), the motion of the blocks is seen as
\begin{align}
x_1(t) =& b_{11} n_1(t), \\
x_2(t) =& b_{21} n_1(t).
\end{align}
If $b_{11} =b_{21}$, then $x_1(t) =x_2(t)$, i.e., the motion of the two blocks are the same. You should be able to see that actually $b_{11} = b_{21}$ for your system.
A: I) Equations of motion
Kinetic energy :
$$T=\frac{m}{2}\left(\dot{x}^2_1+\dot{x}_2^2\right)$$
Potential energy
$$U=\frac{k}{2}\left(x_1^2+(x_2-x_1)^2+x_2^2\right)$$
with Euler Langrage you get:
$${\ddot x}_{{1}}+{\frac {2\,kx_{{1}}-kx_{{2}}}{m}}=0\tag 1$$
$${\ddot x}_{{2}}+{\frac {2\,kx_{{2}}-kx_{{1}}}{m}}=0\tag 2$$
II) Equations of motion:  Normal mode
In normal space  the equations of motion will be:
$$\ddot n_1+\omega_1^2\,n_1=0\tag 3$$
$$\ddot n_2+\omega_2^2\,n_2=0\tag 4$$
to obtain equations (3) and (4)  we have to transformed the coordinates $~x_1~,x_2$ to $~n_1~,n_2$
this can be done with those  equations
$$n_1=\frac 12(x_1+x_2)$$
$$n_2=\frac 12(x_1-x_2)$$
$\Rightarrow~$
$$x_1=n_1+n_2$$
$$x_2=n_1-n_2$$
with this transformation you get:
$$\ddot n_1+\frac km\,n_1=0\tag 5$$
$$\ddot n_2+\frac{3\,k}{m}\,n_2=0\tag 6$$
Remark:
you get the same results equation $(~5~,6~)$if you obtain this transformation :
the center of mass coordinate for n_1:
$$n_1=\frac{m\,(x_1+x_2)}{2\,m}=\frac 12(x_1+x_2)$$
and
$$n_2=\frac 12(x_1-x_2)$$
