There is a fast and down to earth proof. We use the letters $R$ and $L$ to indicate the right and left adjoints of morphisms.

For the first one, just call $a=j_!x\circ y$ and $b=j_!x\circ j_!y$, we have one morphism $\phi:a\to j^*b$ with left adjoint $L\phi:j_!a\to b$. Then your diagram becomes
$$\begin{CD}
j^*j_!a @>j^*L\phi>> j^*b\\
@A\eta AA@|\\
a@>\phi>>j^*b
\end{CD}$$
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 just use adjunction of $j^*$ and $j_!$: your diagram becomes
$$\begin{CD}
j_!j\circ x @>Rj_!(\sigma\ast\text{id})>> j^*j_!x\\
@|@A\sigma\ast\text{id} AA\\
j_!j\circ x@>\text{id}\ast\eta>>j_!j\circ j^*j_!x
\end{CD}$$

We can now unpack
$$Rj_!(\sigma\ast\text{id}):j_!j\circ x\to j^*j_!x$$
as
$$j_!j\circ x\xrightarrow{\sigma\ast\text{id}} x\xrightarrow{\eta} j^*j_!x$$
To check that this unpacking is right, just observe that $R\eta=\text{id}$. Hence the diagram above becomes

$$\begin{CD}
x @>\eta>> j^*j_!x\\
@A\sigma\ast\text{id} AA@A\sigma\ast\text{id} AA\\
j_!j\circ x@>\sigma\ast\eta>>j_!j\circ j^*j_!x
\end{CD}$$
which is obvious.