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?
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:
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$^1$ In particular, this means that OP's interaction term in the Lagrangian should be normalized as $-\frac{\lambda}{4}\color{red}{\phi}^2\color{blue}{\phi}^2$.
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 $$2\times\frac{1}{2!}\times\frac{1}{4^2}=\frac{1}{16}$$
