It would be nice of you to define the involved morphisms in your post. So we consider a diagram with exact rows
$$\require{AMScd}
\begin{CD}
0 @>>> K @>i>> P @>\pi>> M @>>> 0 \\
@. @VV\alpha V @VV\beta V @VV\mathrm{id}_M V @. \\
0 @>>> K' @>>i'> P' @>>\pi'> M @>>> 0
\end{CD}
$$
Here $\beta$ exists by the definition of $P$ being projective, and $\alpha$ is the canonical morphism induced by $\beta$ on the kernels (the restriction to the kernels). Rotman considers then a sequence
$$\tag{*} 0 \to K \xrightarrow{\begin{pmatrix}i\\ \alpha\end{pmatrix}} P\oplus K' \xrightarrow{(\beta, -i')} P' \to 0$$
and claims it is exact. It is really some diagram chasing, but let me write down everything, since you are asking about it.
${i\choose \alpha}$ is injective simply because $i$ is injective.
$(\beta, -i')$ is surjective. For an element $p' \in P'$ we consider $m = \pi' (p')\in M$. Since $\pi$ is surjective, there exists $p\in P$ such that $\pi (p) = m$. We have now
$$\pi' \beta (p) = \pi (p) = m = \pi' (p').$$
That is, $\pi' (\beta (p) - p') = 0$. The kernel of $\pi'$ is $i'$, so there exists some $k'\in K'$ such that $\beta (p) - p' = i' (k')$. Now
$p' = \beta (p) - i' (k')$.
$\operatorname{im} {i\choose \alpha} \subseteq \ker (\beta,-i')$ because the composition of the two arrows is
$$(\beta,-i')\circ {i\choose \alpha} = \beta\circ i - i'\circ \alpha,$$
which is zero by the commutativity of the left square.
Finally, to see that $\ker (\beta,-i') \subseteq \operatorname{im} {i\choose \alpha}$, consider an element $(p,k') \in P\oplus K'$ such that $\beta (p) - i' (k') = 0$. We have by the commutativity of the second square
$$\pi (p) = \pi' \beta (p) = \pi' i' (k') = 0.$$
So $p \in \ker \pi$, i.e. $p = i (k)$ for some $k\in K$. Then, by the commutativity of the first square,
$$i' \alpha (k) = \beta i (k) = \beta (p) = i' (k').$$
Since $i'$ is mono, this implies $k' = \alpha (k)$. So we succeeded in showing that $(p,k') = (i (k), \alpha (k))$ for some $k\in K$.
The exactness of ( * ) implies $K\oplus P'\cong P'\oplus K'$, since $P'$ is projective and hence ( * ) splits.
I checked the proof in Wikipedia that you mention, and it is indeed slightly different: it considers the pullback of the arrows $\pi$ and $\pi'$. Let me write down that argument without using elements.
We consider a pullback square
\begin{CD}
P\times_M P' @>\overline{\pi}>> P' \\
@VV\overline{\pi'} V @VV\pi' V \\
P @>>\pi> M
\end{CD}
It is easy to see that $\overline{\pi'}\colon P\times_M P' \to P$ is an epimorphism with kernel $K'$ and $\overline{\pi}\colon P\times_M P' \to P'$ is an epimorphism with kernel $K$. You can check this yourself from the universal properties of kernels and pullbacks, or look up section 2.5 ("The pullback and pushout theorems") in Freyd's "Abelian categories". Note that pullback of an epimorphism is again an epimorphism, because we work in an abelian category.
So we have short exact sequences
$$0\to K' \to P\times_M P'\ \to P \to 0$$
and
$$0 \to K \to P\times_M P'\ \to P' \to 0$$
Both sequences split since $P$ and $P'$ are projective, hence
$$K'\oplus P \cong P\times_M P' \cong K\oplus P'.$$
