In the following diagram $f,g$ are isomorphisms and $\pi, \tau$ are canonical projections. Is the diagram then commutative ?
$$ \newcommand{\ra}[1]{\kern-1.5ex\xrightarrow{\ \ #1\ \ }\phantom{}\kern-1.5ex} \newcommand{\da}[1]{\left\downarrow{\scriptstyle#1}\vphantom{\displaystyle\int_0^1}\right.} % \begin{array}{ll} A & \ra{g} &B \\ \da{\pi} & &\da{\tau}\\A' & \ra{f} &B' \end{array} $$
($A, B$ are groups) What if $\pi$ and $\tau$ are just surjective ?
No! take: $A,B,A',B'=\mathbb{Z}$ observe that here $\pi$ and $\tau$ are trivial projections. and take $g= id$ and $$f:A' \to B' \\ x\mapsto -x$$ then your claim is wrong since $f\neq g=id$
For your claim to work you need quite precise $f,g$ and quotients, since they heavily depend on each other, also observe that the above example decends to any quotient of $\mathbb{Z}$ with more than $2$ elements.
Also, regarding your last question: every surjection is isomorphic to a quotient, hence that does not change a thing (although this crashes in general either way)