A system is linear if it satisfies two basic principles:
1. Homogeneity (or scaling): If an input \$x_1\$ produces an output \$y_1\$, then an input \$x_2 = \text{K}x_1\$ (where \$\text{K}\$ is a constant) leads to an output \$y_2 = \text{K}y_1\$
2. Additivity If each input \$x_i (i=1, 2,...,n)\$ acting INDIVIDUALLY produces an output \$y_i (i=1, 2,..,\text{n})\$, then an input \$x = x_1 + x_2 + ... + x_n\$ will produce an output \$y = y_1 + y_2 + ... + y_n\$
The two principles can be combined into one, the principle of Superposition:
If each input \$x_i (i=1, 2,...,n)\$ acting INDIVIDUALLY produces an output \$y_i (i=1, 2,.., \text{n})\$, then an input \$x = \text{K}_1x_1 + \text{K}_2x_2 + ... + \text{K}_nx_n\$ will produce an output \$y = \text{K}_1y_1 + \text{K}_2y_2 + ... + \text{K}_ny_n\$, where each \$\text{K}_i (i = 1, 2, ..., \text{n})\$ is a constant.
Therefore:
(A) The system \$v_s = 0.6i_1 - 14v_2\$ satisfies all the principles mentioned above, therefore it is linear. For example, the principle of homogeneity: If the input \$i_1\$ acting alone leads to the output \$v_s'=0.6i_1\$ then the input \$\text{K}i_1\$ acting alone leads to the output \$v_s''=\text{K}0.6i_1\$.
But none of those principles state that, if only \$i_1\$ is multiplied by \$\text{K}\$, the entire expression for \$v_s = 0.6i_1 - 14v_2\$ is multiplied by \$\text{K}\$. So, I think you have mixed up superposition with homogeneity here.
(B) The system \$v_s=0.6i_1v_2\$ is not linear because, when applying additivity, for example, \$i_1=2 \space \text{A}\$, when considered individually, results in an partial output \$v'_s=0 \space \text{V}\$ (since \$v_2=0 \space \text{V}\$, in this case). Similarly, \$v_2=3 \space \text{V}\$, when consideres individually, results in a partial output \$v''_s = \space 0 \space \text{V}\$ (since \$i_1= 0 \space \text{A}\$ in this case). Then, final \$v_s\$ should be \$ 0 \space \text{V}\$, which does not match the real \$v_s=0.6\times 2 \times 3 = 3.6 \space \text{V}\$. For example, if \$v_2\$ were a constant (not an input), something like \$V_2\$, that wouldn't be a problem.
