An odd function is symmetrical in the 1st and 3rd quadrants. Does this means that it always passes through the origin?
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As André Nicolas showed, under your conditions and if $f(0)$ exists, $f(0)=0$. However, nothing in your question implies that $f(0)$ must exist.
If you let $f(x)=\frac1x$ then $f$ is a symmetrical odd function, its graph is in quadrants I and III, but $f(0)$ is undefined.
So, you can say "$f(0)$ is either $0$ or undefined." Or, if you want to stick to terminology about graphs, "the graph of $f$ either passes through the origin or it does not intersect the $y$-axis at all."
Let $f(x)$ be odd, and defined at $x=0$. Then $f(-x)=-f(x)$ for all $x$. In particular, $f(-0)=-f(0)$, so $f(0)=-f(0)$. It follows that $f(0)=0$. The "curve" $y=f(x)$ passes through the origin.