Arbitrary direction of current Kirchhoff's Current Law
You have an error in the "blue arrows" part calculation. Probably it is the sign of the \$2V_0\$ current. The KCL for the junction would be 
$$I_1+I_2-I_3=0$$
while the numbering is from left to right:

\$I_1\$ is the current through the resistor
\$I_2 = 25A\$ 
\$I_3 = -2V_0\$ as it is marked opposite to the actual current source. So the equation becomes:
$$I_1+25+2V_0=I_1+25+2\cdot 10I_1 =0$$
Which leads to 
$$I_1=-25/21=-1.19A$$
So $$V_0=10I_1=-11.9V$$
as expected.