Check my quick proof $P_1 \oplus P_2$ is projective $\iff P_1,P_2$ are projective. Just want to check if I am right in the $\Longrightarrow$ direction. This is Exercise 3 from Dummit Foote. 
($\Longleftarrow$) Given SES $0 \to L \to M \to N \to 0$, $ Hom(P_1, -) \oplus Hom(P_2,-) = Hom(P_1 \oplus P_2, -)$. Take the direct sum of the two SES of $Hom(P_1, -)$ and $Hom(P_2, -)$ to get this direction.
($\Longrightarrow$) 
WLOG, taking $P_1$. I am using Dummit&Foote Proposition 30.2 (one of the projective equivalences). The lowest row is exact (right-exact)
$$\begin{array}{} P_1 & \xrightarrow{g} & P_1 \oplus P_2 & \\
& \swarrow{F} & \downarrow{f} &  \\
M & \xrightarrow{\phi} & N & \xrightarrow{} & 0 & \end{array}$$
Then defining $F' = F\circ g$ and $f' = f\circ g$ are the require homomorphisms.
I forget to mention, $g: P_1 \to P_1 \oplus P_2$ is just $p_1 \to (p_1, p_2)$

To show $P_1$ is projective you need to show that any map $f:P_1 \to N$ factors through M, that is, show there exists a map $g: P_1 \to M$ such that $\phi g = f$. So your idea to involve the direct sum is right but you need to switch up the order of operations a bit.
An alternative would be to use the fact that a module is projective if and only if it's a direct summand of a free module. You can use this to do both directions actually.
