Let $x = u(t)$ and $y = v(t)$ be a parametric representation of a curve.
If we assume $u$ is one-one on some open interval in $t$, we can define $f = v \circ u^{-1}$. Then the curve $(u(t), v(t))$ on that interval is the curve $(x, f(x))$ for $x = u(t)$.
Next if we assume $u$ and $v$ are differentiable and $u'(t) \neq 0$, then we have
$$f'(x) = \frac{v'(t)}{u'(t)}. \tag{*}\label{*}$$
Usually I see the slope of the tangent line defined by \eqref{*}. But what happens if u is not one-one on any interval surrounding $t$ and $u'(t) \neq 0$? For example
$$u(t) = \begin{cases} t^2 \sin\left(\frac{1}{t}\right) + \frac{t}{2} & t \neq 0 \\ 0 & t = 0 \end{cases}$$
at $t = 0$. We can still compute $v'(t)/u'(t)$ and it seems like a reasonable slope for the tangent, but we cannot define an $f$. So it seems more straightforward to define the slope of the tangent line to be $v'(t)/u'(t)$. Then we have a theorem that says that $f'(x) = u'(t)/v'(t)$ when $f(x)$ exists.