Intuition regarding linear combinations of conditioned probabilities I need your help.
Let $X,Y$ denote real-valued bounded random variables. Assume that the conditioned probability $P(X|Y)$ can be written as a linear combination of $N=3$ real-valued functions $f_i(X)$, i.e. $P(X|Y) = \sum_{i=1}^N c_i(Y) f_i(X)$. The coefficients $c_i$ and functions $f_i$ could be negative.
Then, can I always find exlusively non-negative coefficients $c'_i$ and non-negative functions $f'_i$ so that $P(X|Y) = \sum_{i=1}^N c'_i(Y) f'_i(X)$?
For three different $P(X|Y_i)$ I could span $P(X|Y) = \sum_{i=1}^N c'_i(Y) P(X|Y_i)$, but could I choose those $P(X|Y_i)$ such that $c'_i(Y) \ge 0$? Maybe if they are within the convex hull?
 A: If I understand you correctly, the conjecture is false for $N=3$ and here is a counter-example.  [Partial credit to joriki whose deleted (wrong) answer led me down the eventually correct path.]
Suppose $X$ takes only four different values, and same for $Y$.  So $g_i(x) = P(X = x \mid Y = y_i)$ is just a $4$-vector $\in S = \{(a,b,c,d) \in \mathbb{R}^4: a+b+c+d = 1 \cap a,b,c,d \ge 0\}$.
Here are the four $4$-vectors for $g_1, g_2, g_3, g_4$:  
$$(\frac12, \frac12, 0, 0),\,\,\,\,\,\, (0, \frac12, \frac12, 0),\,\,\,\,\,\, (0, 0, \frac12, \frac12),\,\,\,\,\,\, (\frac12, 0, 0, \frac12)$$
Since the vectors are linearly dependent (with any three of them acting as a basis), this satisfies the pre-condition.  
Proving that the post-condition is impossible is a harder, and I can only present an intuitive argument.  Any candidate $f'_i$ with three or four positive entries are useless - the coefficients $c'_i$ are non-negative so such an $f'_i$ cannot be used in a positive linear combination (plc) of any $g_i$ which has only two positive entries.  Similarly, any candidate $f'_i$ with two positive entries (and two zeros) can only be used in a plc of the "matching" $g_i$, i.e. one with the same two entries being positive.  If we take out such an $f'_i$ and the "matching" $g_i$ (if any), the problem reduces to using one fewer $f'_i$ to represent (via plc) maybe one fewer $g_i$.  After we eliminate all $f'_i$'s with two positive entries (and any matching $g_i$'s), we are left with $K$ number of $f'_i$'s, each having only one positive entry, and requiring them to  represent (via plc) at least $K+1$ remaining $g_i$'s.  A simple check of $K=0,1,2,3$ shows that this is impossible.
Sorry my argument is so informal!  I had a hard time writing it out algebraically.

Further remarks: The key goemetric reason for the impossibility is that $S$ is a tetrahedron, which has a cross-section which is a square, and indeed $g_i$'s are the corners.  This almost immediately shows that using $f''_i$ which are $\in S$ (i.e. they are pdf's) and convex coefficients $c''_i$ does not suffice.  It takes a bit more work (above) to show the more general case that using non-negative $f'_i$ and non-negative coefficients $c'_i$ also does not suffice.
Curiously, similar logic would say the conjecture is true for $N=2$.  In this case, linear combinations of $f_1, f_2$ is a line, whose intersection with $S$ (any $S$, not just this tetrahedron case) is one or more line segments.  If we pick the two furthest points of these line segments (the limits exist since $X$ is bounded) as $f'_1, f'_2$, then everything can be represented as plc.
