The definition of a bounded linear operator is a linear transformation $T$ between two normed vectors spaces $X$ and $Y$ such that the ratio of the norm of $T(v)$ to that of $v$ is bounded by the same number, over all non-zero vectors in $X$.
What is this definition saying, is it saying that the norm of vectors is preserved under the transformation?
Why does ratio come into it?, the ratio of what?
On
What the definition is saying is that there exists a constant $C>0$ such that, for every $v\in X$: $$\|T(v)\|\le C\|v\|.$$ In other words, the ratio $\frac{\|T(v)\|}{\|v\|}$ is bounded by the same constant $C$, regardless $v\in X$. Note that the latter requires $v\ne 0$ for the ratio to be defined.
Let $B_r(x)$ denote the open ball of radius $r$ centered at $x$: $$ B_{r}(x) = \{ y : \|x-y\| < r \}. $$ A linear transformation $T$ is continuous at $x$ iff, for every $\epsilon > 0$ there exists $\delta > 0$ such that $$ T(B_{\delta}(x)) \subseteq B_{\epsilon}(Tx) $$ Because of linearity, the above is equivalent to $$ T(B_{\delta}(0))\subseteq B_{\epsilon}(0). $$ So $T$ is continuous at $0$ iff it is continuous at every $x$. Furthermore, because of how linear transformations commute with scalar multiplication, continuity at $0$ is equivalent to the existence of $\delta > 0$ such that $$ T(B_{\delta}(0)) \subseteq B_{1}(0). $$ Indeed the above holds for some $\delta > 0$ iff $$ T(B_{\delta\epsilon}(0))\subseteq B_{\epsilon}(0). $$ Another equivalent is that $T(B_{1}(0))$ is contained in some finite ball $B_{r}(0)$. That is, $T$ is continuous at every $x$ iff there exists $r > 0$ such that $$ T(B_{1}(0)) \subseteq B_{r}(0). $$ Equivalently, there exists $r > 0$ such that $$ \|Tx\|_{Y} \le r\|x\|_{X},\;\;\; x\in X. $$ This last condition is the definition of a bounded transformation.