Let $0\to L\stackrel{\alpha}\to M\stackrel{\beta}\to N\to 0$ be a splitting s.e.s. where $\alpha$ has a retraction $r$.
(a) Show that in this case $M=\alpha(L)\oplus\ker(r)$.
I am a little bit confused because we have defined $\oplus$ as "direct sum" that is $A\oplus B$ is all ordered pairs $(a,b)$ with $a\in A$ and $b\in B$. This cannot be equal to $M$ (it can at most be homomorphic to $M$). How can I otherwise interpret $\oplus$?
The notation $\oplus$ is used in two slightly different meanings. If $A$ and $B$ are submodules of $M$, then $A+B$ is written $A\oplus B$ when $A\cap B=\{0\}$. If $M$ and $N$ are modules, then $M\oplus N$ denotes the module whose elements are ordered pairs, with componentwise addition and multiplication by scalars.
The direct sum $M\oplus N$ (in the second sense) can be seen as $M'\oplus N'$ in the first sense, where $M'=\{(x,0):x\in M\}$ and $N=\{(0,y):y\in N\}$. Since $M$ and $M'$ are “canonically” isomorphic by $x\mapsto(x,0)$ and similarly for $N$, using the same notation is abusive, but handy. Often the two are distinguished by saying “internal direct sum” (first sense) and “external direct sum” (second sense).
What you have to prove is that the obvious morphism $\sigma$ from $\alpha(L)\oplus\ker r$ (external) to $M$ defined by $$ \sigma\colon (x,y)\mapsto x+y $$ is an isomorphism. Since $\ker\sigma=\{(x,y):y=-x\}$, this amounts to saying that
which is the same as saying that $\alpha(L)\oplus\ker r$ (internal direct sum) equals $M$.
So, let $z\in\alpha(L)\cap\ker r$; then $z=\alpha(x)$ for some $x\in L$, so $$ 0=r(z)=r\alpha(x)=x $$ because $r$ is a retract of $\alpha$. Therefore $z=\alpha(0)=0$.
For the second condition, let $z\in M$; then $x=r(z)\in L$ and, setting $y=z-\alpha r(z)=z-\alpha(x)$ we have $$ r(y)=r(z-\alpha r(z))=r(z)-r\alpha r(z)=r(z)-r(z)=0 $$ so $y\in\ker r$. Therefore $z=\alpha(x)+y\in\alpha(L)+\ker r$.