Now the definition of an independent loop is a loop that contains a branch that is not part of any other independent loop.
If a loop has a branch that's not part of any other loop, that does guarantee independence, but I don't think it's required. (Mathematically, it's sufficient but not necessary.)
In mesh analysis, you're trying to solve a system of equations. For that, you need one equation per variable. But the equations must be linearly independent -- if you can make one equation by adding, subtracting, and/or multiplying the other equations, it doesn't count. For example:
$$x + y = 5$$
$$2x + 2y = 10$$
The second equation can be produced by doubling every value in the first equation. This doesn't give you any new information, so you can't solve for x and y. But in this example:
$$x + y = 5$$
$$x + 2y = 7$$
you can't get the second equation by manipulating the first. So you can find the solution: x = 3 and y = 2.
Back to circuits. Your system has three variables -- the mesh currents \$I_L\$ (on the left), \$I_M\$ (in the middle), and \$I_R\$ (on the right). Here are the equations, assuming the mesh currents flow clockwise:
$$10\mathrm V - I_L \cdot 5 \Omega - (I_L - I_M) \cdot 2 \Omega = 0$$
$$-(I_M - I_L) \cdot 2 \Omega - (I_M - I_R) \cdot 3 \Omega = 0$$
$$I_R = -2\mathrm A$$
Grouping the variables gives:
$$10 \mathrm V - I_L \cdot 7 \Omega + I_M \cdot 2 \Omega = 0$$
$$I_L \cdot 2 \Omega -I_M \cdot 5 \Omega + I_R \cdot 3 \Omega= 0$$
$$I_R = -2 \mathrm A$$
There's no way we can make one of these equations out of the other. The first has a constant term, the second doesn't, and the third just gives us the value of one variable. If we substitute \$-2\mathrm A\$ for \$I_R\$ and try to make the signs match, it's even more obvious:
$$I_L \cdot 7 \Omega - I_M \cdot 2 \Omega - 10 \mathrm V = 0$$
$$I_L \cdot 2 \Omega - I_M \cdot 5 \Omega - 6 \mathrm V = 0$$
The ratios of the coefficients and constants are totally different. These equations are linearly independent.