I have two observations, which stop short of a full
solution.

First, I have noticed that there can be no such sequence of
functions $f_n$, if one insists that every $f_n$ is a
measurable function. In particular, there can be no squence of Borel functions with your property. To see this, suppose that
$f_n:\mathbb{R}\to\mathbb{R}$ is a countable sequence of
measurable functions. If infinitely many $f_n$ are not
surjective, then clearly the property will fail and we are
done. So we may assume that all but finitely many of the
functions are surjective, and by throwing away finitely
many functions, we reduce to the case where all $f_n$ are surjective.
For each natural number $n$ and real $y$, let
$A^y_n=f_n^{-1}(y)=\{\ x\ \mid\ f_n(x)=y\ \}$. For any fixed
$n$, the sets $A^y_n$ for various $y$ partition
$\mathbb{R}$ into continuum many disjoint measurable sets.
Since one cannot have uncountably many disjoint positive
measure sets, it follows that there must be some $y_n$ (and
in fact many such $y_n$) with $\lambda(A^{y_n}_n)=0$, where
$\lambda$ is the Lebesgue measure. Let
$K=\mathbb{R}-\bigcup_n A^{y_n}_n$, which is the complement
of a measure $0$ set, and so in particular, $K$ has size
continuum. Observe that $f_n[K]$ omits $y_n$ by
construction, and so there is no $n$ for which
$f_n[K]=\mathbb{R}$, showing that the desired property
fails. I guess the argument actually doesn't need that
every $f_n$ is measurable, but only that $f_n^{-1}(y)$ is a
measurable set for every $y$.

My second observation is that I noticed that if the Continuum Hypothesis holds,
then there is a sort-of-near-positive solution in the sense that there is a sequence of functions
$f_n:\mathbb{R}\to\mathbb{R}$ such that for every
uncountable set $K\subset\mathbb{R}$, the *combined* range is
onto, meaning $\bigcup_n f_n[K]=\mathbb{R}$. (And in fact,
the existence of such $f_n$ is equivalent to CH.) To see
this, let $\lt$ well-order $\mathbb{R}$ in order type
$\omega_1$. In particular, every real $x$ has only
countable many predecessors in $\lt$, so enumerate them
as $f_n(x)$ for $n\in\omega$. If $K\subset\mathbb{R}$ is
any uncountable set, then $K$ is unbounded in the
well-order, and so every real $y$ is a predecessor of some
element of $K$, and so $y=f_n(x)$ for some $x\in K$. Thus,
$\bigcup_n f_n[K]=\mathbb{R}$, as desired.

I don't know the answer to the fully general question you
asked.