Let $f(x)=ax^{2}+bx+c(a≠0)$, then $f(x)+g(x)=(a-1)x^{2}+bx+c-3$,

Since $f(x)+g(x)$ is an odd function, $a=1$, $c=3$

Thus, $f(x)=x^{2}+bx+3$, symmetry axis $x=- \frac {b}{2}$,

$\mathbf{1^{st}}$ case: When $- \frac {b}{2} > 2$, i.e., $b 2$, $f(x)$ is an increasing function on $[-1,2]$,

Thus, the minimum value of $f(x)$ is $f(-1)=4-b=1$,

Thus, $b=3$, in this case $f(x)=x^{2}+3x+3$,

In summary, $f(x)=x^{2}-2 \sqrt {2}x+3$, or $f(x)=x^{2}+3x+3$.

Hence, the final answers are $\boxed{f(x)=x^{2}-2 \sqrt {2}x+3}$ and $\boxed{f(x)=x^{2}+3x+3}$.