Symmetry factors in two interacting fields 
Red and blue colored lines represent the two different fields.
At 1st order, by the exchange of the blue legs and red legs we get $\frac{1}{4}$ factor and in one of the 2nd order term drawn above, from the exchange of vertices, the blue legs and red legs we get $\frac{1}{2^4 \cdot 2}$.
However it seems like I'm missing a factor of 2 in the numerator for the number of topologically different feynman diagrams.
Could anyone point out what I'm missing here?
 A: If we are using the standard convention, where each term$^1$ in the Lagrangian is divided by its symmetry factor, then the numerical coefficient in front of a Feynman diagram is the reciprocal of its symmetry factor $S$.
For a proof, see e.g. Ref. 1.
OP's first diagram has $S=4$ while OP's second diagram has $S=16$.
References:

*

*P. Etingof, Geometry & QFT, MIT 2002 online lecture notes; Chapter 3.

--
$^1$ In particular, this means that OP's interaction term in the Lagrangian should be normalized as $-\frac{\lambda}{4}\color{red}{\phi}^2\color{blue}{\phi}^2$.
A: Assuming we have the interaction $$\mathcal{L}_{\text{int}}=\frac{\lambda}{2\times2}\phi^2\Phi^2$$
The first bubble diagram comes from the $1$st-order expansion of the expression
$$\langle \Omega|Te^{-i\int d^4x\mathcal{L}_{\text{int}}}|\Omega\rangle$$
which is
$$\frac{-i\lambda}{4}\int d^4x\langle \Omega|T\phi(x)^2\Phi(x)^2|\Omega\rangle$$
There's only one way to contract this expression, namely $\phi$ with $\phi$ and $\Phi$ with $\Phi$, and so we just have the $1/4$.
The next two-loop bubble diagram comes from the $2$nd-order expansion
$$\frac{1}{2!}\left(\frac{-i\lambda}{4}\right)^2\int d^3x ~d^3y \langle \Omega|T\phi(x)^2\Phi(x)^2\phi(y)^2\Phi(y)^2|\Omega\rangle$$
There's a bunch of terms in this expression, the one we want is generated by contracting the 2 $\Phi(x)$s and $\Phi(y)$s together. There are two ways for the 4 $\Phi$s to contract with each other and only one way for the remaining $\phi$s to contract at the same points. And so the overall numerical factor in front of our integral here is
$$2\times\frac{1}{2!}\times\frac{1}{4^2}=\frac{1}{16}$$
