Direct sums in projective modules

Since $P_1$ is a direct summand, there exists $P_2$ such that $P=P_1\oplus P_2$. Consider $f:P\rightarrow P_2$ such that for every $x\in P$, write $x=x_1+x_2, x_1\in P_1,x_2\in P_2$, $f(x)=x_2$. Let $x\in P_2$, we can write $x=x_1'+n, x_1'\in P_1, n\in N$, we have $x=f(x)=f(x_1'+n)=f(n)$. This implies that the restriction $g$ of $f$ to $N$ is surjective, there exists $h:P\rightarrow N$ such that $f=g\circ h$; $P=P_1\oplus h(P_2)$.

Let $x\in P$, write $x=x_1+x_2, x_1\in P_1, x_2\in P_2$, $f(x-h(x_2))=f(x)-f(h(x_2))=x_2-x_2=0$ since $x_2=f(x_2)=f(h(x_2))$. This implies that $x-h(x_2)\in P_1$, we can write $x=x-h(x_2)+h(x_2)$ and we deduce that $P=P_1+h(P_2)$.

Let $x\in P_1\cap h(P_2)$, we can write $x=h(x_2), x_2\in P_2$, $f(x)=f(h(x_2))=f(x_2)=x_2=0$ because $x\in P_1$. This implies that $x_2=x=0$.