Can $ f(c | a, b) \propto f(a | b, c) $? When? Given that
$$\begin{align}
\mathsf f(a\mid b, c) & = \frac{\mathsf f(a,b,c)}{\mathsf f(b, c)}
\\[1ex] & =\frac{\mathsf f(c\mid a,b)\,\mathsf f(a, b)}{\mathsf f(b, c)}
\\[1ex] & =\frac{\mathsf f(c\mid a,b)\,\mathsf f(a\mid b)\,\mathsf f(b)}{\mathsf f(b, c)}
\\[1ex] & =\frac{\mathsf f(c\mid a,b)\,\mathsf f(a\mid b)}{\mathsf f(c\mid b)}\frac{\mathsf f(b)}{\mathsf f(b)}
\\[1ex] & =\frac{\mathsf f(c\mid a,b)\;\mathsf f(a\mid b)}{\mathsf f(c\mid b)}
\end{align}$$
and assuming that there is no prior information about $a$, $b$ and $c$ and $a, b$ are mutually independent, then $ f(a\mid b) = f(a) = constant $.
Hence, can't I assume that $ f(a | b, c) \propto f(c | a, b) $?

In general, if $a$ and $b$ are independent, you have:
$$\begin{align}
f(c|a,b)
&= \frac{f(a,b,c)}{f(a,b)} \\[6pt]
&= \frac{f(a,b,c)}{f(b,c)} \cdot \frac{f(b,c)}{f(a,b)} \\[6pt]
&= f(a|b,c) \cdot \frac{f(b,c)}{f(a) f(b)} \\[6pt]
&= f(a|b,c) \cdot \frac{f(c|b)}{f(a)}. \\[6pt]
\end{align}$$
So, under this condition you have the logical equivalence:
$$f(c|a,b) = f(a|b,c)
\quad \quad \quad \iff \quad \quad \quad 
f(c|b) = f(a).$$
In other words, under your independence assumption, you can still only say that $f(c|a,b) = f(a|b,c)$ if $f(c|b) = f(a)$.
