Automorphism of $P\oplus P'$ that induces isomorphism $P\oplus K'\cong P'\oplus K$. 
Let $P,P'$ be two projective $R$-modules, and $K\subset P,K'\subset P'$ be submodules  such that $P/K\cong P'/K'$. Show that there exists an automorphism $\phi$ of $P\oplus P'$ such that $\phi(P\oplus K')=P'\oplus K$. In particular, $P\oplus K'\cong P'\oplus K$.

It is not difficult to show directly that $P\oplus K'\cong P'\oplus K$, for instance, by taking the pullback square of $P\to P/K, P'\to P'/K'\cong P/K$. But I don't know how to construct such a $"\phi"$ in the question. Any help?
 A: Just recall the classical definition in the process of proving Schanuel's lemma
$$ X=\{ (p,p')\in P\oplus P' :\pi(p)=\pi '(p') \} \; .$$
And try to write down explicit maps how $X$ isomorphic to $P\oplus K'$ and $K\oplus P'$  using two lifted maps $P\rightarrow P'$ and $P'\rightarrow P$ compatible with the projection.
A: Special case:
Assuming $P,P'$ are vector spaces and $R$ is a field. Note that this is a very special case of projective R modules.
Let $P = <v_1,...,v_{\ell_p},k_1,...,k_{a_p}>$ where $K = <k_1,...,k_{a_p}>$. 
Let $P' = <w_1,...,w_{\ell_{p'}},k'_1,...,k'_{a_{p'}}>$ where $K' = <k'_1,...,k'_{a_{p'}}>$.
Since $P/K \tilde{=} P'/K'$, we have $\ell_p = \ell_{p'} = \ell$.
Let $a_{p} \geq a_{p'}$.
Define $$\phi(<v_1,...,v_{\ell}>)=<w_1,...,w_{\ell}>$$
$$\phi(<w_1,...,w_{\ell}>)=<v_1,...,v_{\ell}>$$
$$\phi(<k_{a_{p'}+1},...,k_{a_p}>) = <k_{a_{p'}+1},...,k_{a_p}>$$
$$\phi(<k_{1},...,k_{a_{p'}}>) = <k'_{1},...,k'_{a_{p'}}>$$
$$\phi(<k'_{1},...,k'_{a_{p'}}>) = <k_{1},...,k_{a_{p'}}>$$
This is the desired automorphism $\phi$.
Try generalizing this to projective R modules. I am aware that this is a proof of special case. But hope that this helps.
The result you want is true if you can follow/prove the steps below in a given module:

*

*Try proving $P = P/K \oplus K$, $P' = P'/K' \oplus K'$.

*Try forming a surjective homomorphim between $K$ and $K'$ or viceversa.

*Write for example $K \tilde{=} M \oplus N$ with $M \tilde{=} K'$ using the above surjective morphism

*Define $\phi$ by:
$$\phi(P/K) = P'/K'$$
$$\phi(P'/K') = P/K$$
$$\phi(N) = N$$
$$\phi(M) = K'$$
$$\phi(K') = M$$
