Let $B_t$ be a standard BM and $M_t := \int_0^t e^{B_s} \, dB_s$. The task is to compute the expectation
\begin{align*} \mathbb{E}(M_1 M_2). \end{align*}
Now is it legal to calculate, by using $dM_t = e^{B_t} \, dB_t$, that
\begin{align*} E(M_1M_2) &= E \biggl( \int_0^2 \int_0^1 e^{B_t}e^{B_s} \, dB_s dB_t \biggr) \\ &= E \biggl( \int_0^2 \underbrace{\int_0^1 1 \, dM_s}_{\stackrel{d}{=} \mathcal{N}(0,1)} dM_t \biggr) = \frac{1}{\sqrt{2}} E(XY) = \frac{1}{\sqrt{2}} E(X) E(Y) = 0, \end{align*}
where $X$ and $Y$ are independent standard normal?
Hint
There is more or less nothing true in all what you said. You don't have Fubini with Itô integrals. Moreover, it's not clear what are $X$ and $Y$ in your question. Finally, it's not true that $M_1\sim \mathcal N(0,1)$.
Let $(\mathcal F_t$) being the natural filtration of $(B_t)$. Using basic properties of Itô-integral yields,
\begin{align*} \mathbb E\left[\int_0^1e^{B_s}\,\mathrm d B_s\int_0^2e^{B_s}\,\mathrm d B_s\right]&=\mathbb E\left[M_1^2\right]+\mathbb E\left[\int_0^1e^{B_s}\,\mathrm d B_s\mathbb E\left[\int_1^2e^{B_s}\,\mathrm d B_s\mid \mathcal F_1\right]\right]\\ &=\mathbb E[M_1^2]+\mathbb E\left[\int_0^1e^{B_s}\,\mathrm d B_s\right]\mathbb E\left[\int_1^2e^{B_s}\,\mathrm d B_s\right]. \end{align*}
I let you conclude.