Can I apply the Fundamental Theorem of Calculus for $$\int_{-\infty}^{t_1} \int_{-\infty}^{t_2} \frac{\partial \phi\left(\frac{z_2 - \rho z_1}{\sqrt{1 - \rho^2}}\right)}{\partial z_2} dz_2 dz_1$$ in order to get $$\int_{-\infty}^{t_1} \phi\left(\frac{t_2 - \rho z_1}{\sqrt{1 - \rho^2}}\right)dz_1,$$ where $\phi(\cdot)$ is standard normal pdf?
I am not sure whether I can treat $z_1$ as fixed and take $\lim_{a \rightarrow -\infty} \phi\left(\frac{a - \rho z_1}{\sqrt{1 - \rho^2}}\right)$.
Yes. You can write $$\int_{a}^{b}\frac{\partial \phi(z_{1},z_{2})}{\partial z_{2}}d z_{1}=\frac{d}{d z_{2}}\int_{a}^{b}\phi(z_{1},z_{2})dz_{1}$$
Assuming $t_{1},t_{2}$ are constants, we have $$\int_{-\infty}^{t_{2}} \int_{-\infty}^{t_{1}}\frac{\partial \phi(u(z_{1},z_{2}))}{\partial z_{2}}dz_{1}dz_{2}=\int_{-\infty}^{t_{1}}\phi(u(t_{2},z_{1}))dz_{1}-\int_{-\infty}^{t_{1}}\lim _{x \to -\infty}\phi((u(x,z_{1}))dz_{1}$$ where we use various continuity assumptions to move the limit around. Now, $$\lim_{x \to -\infty}\frac{x-\rho z_{1}}{\sqrt{1-\rho^{2}}}=-\infty$$, so as $x \to -\infty$, $u \to -\infty$ and $\phi(u) \to 0$