"( foo && bar && baz ); finally-code" <- how can finally-code not be run? - node.js

Question

I have this node.js code which generates a bash script:
const k = cp.spawn('bash');
let cmd = [
`cd ${v.path}`,
`git checkout "${releaseName}"`,
`git add .`,
`git commit -am "modified package.json"`,
`git checkout master`,
`( git branch -D -f master_copy_npp_tool &> /dev/null || echo "" ) `,
`git checkout -b master_copy_npp_tool`,
`git merge --no-commit -m "This release branch should be merged into master and integration." "${releaseName}"`,
`git checkout ${releaseName}`,
`git push -u origin ${releaseName}`
]
.join(' && ');
cmd = `( ${cmd} ); git checkout -f master; `; // always checkout the integration branch again, at the end
k.stdin.end(cmd);
For some reason, the script leaves the git repo on the releaseName branch, not master.
The command as a string looks like:
( cd /home/rolo/foo
&& git checkout "oleg/npp_tool/release/1531621240127"
&& git add . && git commit -am "modified package.json"
&& git checkout master
&& ( git branch -D -f master_copy_npp_tool &> /dev/null || echo "" )
&& git checkout -b master_copy_npp_tool
&& git merge --no-commit -m "(shortened)." "oleg/npp_tool/release/1531621240127"
&& git checkout oleg/npp_tool/release/1531621240127
&& git push -u origin oleg/npp_tool/release/1531621240127 );
git checkout -f master;
I am trying to use the last clause:
git checkout -f master;
as a "finally" block. I want to always checkout the master branch in the end no matter way. Anyone know what's going wrong? Like I said it remains on the release branch.
Basically what I tried to do is this pattern:
( foo && bar && star ); finally-clause;