Bounded linear transformation definition clarification

Clearly, if there is an $M > 0$ such that $\frac{\|Tx\|_Y}{\|x\|_X}\leq M$ for all $x\in X$, then $T$ maps bounded sets to bounded sets. Take $S\subset X$ such that $\|x\|_X\leq B < \infty$ for all $x\in S$. Then, for $y\in T(S)$, $y = Tx$ for some $x\in S$, which implies that $\|y\|_Y\leq M\|x\|_X\leq MB$, so $T(S)$ is bounded.

Similarly, if $T$ maps bounded sets to bounded sets, then we must have some $M\geq 0$ such that $\frac{\|Tx\|_Y}{\|x\|_X}\leq M$. Otherwise, take an increasing sequence $\{M_n\}_{n=1}^{\infty}$ such that $M_n\to \infty$, and for each $n$ choose an $x_n\in X$ such that $\|x_n\|_X = 1$ and $\frac{\|Tx_n\|_Y}{\|x_n\|_X} = \|Tx_n\|_Y > M_n$, which we can do by the linearity of $T$, as $$\frac{\|Tx\|_Y}{\|x\|_X} = \left\|T\frac{x}{\|x\|_X}\right\|_Y$$ is only dependent on the "direction" of $x$, not on its length. Then, $\{x_n\}_{n=1}^{\infty}$ is bounded as it lies within the unit sphere, but $\{Tx_n\}_{n=1}^{\infty}$ is not bounded.

The condition that you have written is the same as $$ \sup_{||x||\leq1} ||Tx||_Y = \sup_{x\in X} \frac{||Tx||_Y}{||x||_X}\leq M$$ because $T $ is linear and $||\lambda x|| = |\lambda|||x||$.

Therefore, if $T$ sends a bounded sets to bounded sets it sends the unit ball in a bounded set (so $M$ exists). Vice versa, if the inequality above holds, take a bounded set $K$, it must be contained in a ball or radius $r>0$, i.e. $K \subset r B_X $, it follows that $T(K)\subset M r B_Y$ (where $B_X$ is the unit ball in the space $X$).