= Solution
Take the conducting plane as $y=0$, with the real <point charges> $Q$ at $(0,b)$ and $-2Q$ at $(a,b)$. The <method of images> replaces the plane by image charges $-Q$ at $(0,-b)$ and $+2Q$ at $(a,-b)$. This makes the plane an equipotential; the physical field above it is the sum of the real- and image-charge fields. No factor of $1/2$ enters the force on a real charge.
Set $k_e=1/(4\pi\epsilon_0)$ and $d=\sqrt{a^2+4b^2}$, and use positive $x$ from $Q$ toward $-2Q$, positive $y$ upward. The horizontal forces combine the mutual attraction and the force from the other charge's image:
$$
\boxed{F_{Q,x}=2k_eQ^2\left(\frac1{a^2}-\frac a{d^3}\right),\qquad F_{-2Q,x}=-F_{Q,x}.}
$$
For the vertical components, the own-image attractions act downward; each cross-image term acts upward. Thus
$$
\boxed{F_{Q,y}=k_eQ^2\left(-\frac1{4b^2}+\frac{4b}{d^3}\right),\qquad F_{-2Q,y}=k_eQ^2\left(-\frac1{b^2}+\frac{4b}{d^3}\right).}
$$
These are the electrostatic forces; the external forces holding the charges fixed are their negatives. The smaller positive charge's vertical force need not always point downward, because the other charge's induced surface charge also contributes. The construction assumes the usual ideal large-plane limit with no imposed background field.
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