Transfer Function definition
We do not define it in that manner. Note the difference between the time and Laplace domains.

Generally, we characterize a system by the response produced when the system is excited by an impulse response $\delta(t)$. Given the output $y(t)$ for such signal we ask which function $h(t)$ did we convolve with the $\delta(t)$ to produce it. Formulation:
$$\delta(t)*h(t)=y(t)$$

Here $*$ represent convolution and not a product. When we find $h(t)$ we can produce the output for any other input $x(t)$ by means of convolution:
$$x(t)*h(t)=y(t)$$

If we transform into the Laplace domain, a convolution is translated into multiplication:

$$\mathcal{F}\{x(t)*h(t)\}=\mathcal{F}\{y(t)\}$$
$$\mathcal{F}\{x(t)\}\cdot \mathcal{F}\{h(t)\}=\mathcal{F}\{y(t)\}$$
$$X(s)\cdot H(s)=Y(s)$$
$$H(s)=\frac{Y(s)}{X(s)}$$

It is a convention that the capital letters represent the transformed signals.  Also, $t$ has changed into $s$ because we are now in the Laplace domain. Since the convolution transformed into a product we may now divide both sides by $X(s)$. We could not do this on the time domain because there we had a convolution operator and not a product.