For the first one, just call $a=j_!x\circ y$ and $b=j_!x\circ j_!y\circ j$, we have one morphism $a\to b$. Then your diagram becomes
$$\begin{CD}
j^*j_!a @>>> b\\
@VVV@|\\
a@>>>b
\end{CD}$$
which is an easy $1$ categorical fact about adjunction.

For the second one, observe that you can prove commutativity before composing with $y$: every arrow has the form $\phi\ast y$ for some $\phi$. Now observe that $j_!(j_!j\circ x)\to j_!j\circ j_!x$ has a left inverse induced by $j_!j\circ j^*j_!x\to j_!j\circ x$. The fact that this is a left inverse is precisely the first diagram with substitutions $x\mapsto j$, $y\mapsto x$. By plugging the left inverse in the second diagram, we reduce ourselves to prove the commutativity of
$$\begin{CD}
j_!j\circ j_!x@>>>j_!(j_!j\circ x)\\
@VVV@VVV\\
j_!x@=j_!x
\end{CD}$$ 
Now, using the adjunction between $j_!$ and $j^*$, the diagram above is equivalent to
$$\begin{CD}
j_!j\circ j_!x\circ j@>>>j_!j\circ x\\
@VVV@VVV\\
j_!x\circ j@>>>x
\end{CD}$$
which is obvious, it's just $a\ast b=(a\ast\text{id})\circ (\text{id}\ast b)=(\text{id}\ast b)\circ (a\ast\text{id})$.