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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.

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. It's just the right adjoint diagram of $L\phi\circ\text{id}=\text{id}\circ L\phi$, with $\eta=R\text{id}$ and $\phi=RL\phi$.

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.

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.

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.

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 $a\ast b=(a\ast\text{id})\circ (\text{id}\ast b)=(\text{id}\ast b)\circ (a\ast\text{id})$.

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.