For the first question, I will assume that $f(x,y)\to 0$ when $x^2+y^2\to\infty$. Since $$I(x,y,z)=\iint_{\mathbb R^2}\frac z{2\pi(t_1^2+t_2^2+z^2)^{\frac 32}}f(t_1+x,t_2+y)dt_1dt_2,$$
and $f$ is bounded, for a fixed $\varepsilon>0$, we can find $R>0$ such that
$$\left|\iint_{t_1^2+t_2^2\geq R^2}\frac z{2\pi(t_1^2+t_2^2+z^2)^{\frac 32}}f(t_1+x,t_2+y)dt_1dt_2\right|\leq\varepsilon.$$
Since for $t_1^2+t_2^2\leq R$ we have $||(t_1+x,t_2+y)|| \geq \sqrt{x^2+y^2}-\sqrt{t_1^2+t_2^2}\geq \sqrt{x^2+y^2}-R,$
and we can find $R'$ such that if $a^2+b^2\geq R'$ then $|f(a,b)|\leq\varepsilon$, if $x^2+y^2\geq R+R'$ we have $|I(x,y,z)|\leq\varepsilon$.
For the second question, fix $\varepsilon>0$. Take $R>0$ such that
$$\left|\iint_{t_1^2+t_2^2\geq R^2}\frac z{2\pi(t_1^2+t_2^2+z^2)^{\frac 32}}dt_1dt_2\right|\leq\varepsilon.$$
(note that such a $R$ doesn't depend on $z$ since it's the case for $\int_{\mathbb R^2}\frac{z}{(u^2+v^2+z^2)}dudv$)
Then we use the uniform continuity of $f$ on $\overline{C(0,R)}$, we take $\delta>0$ such that if $x_1^2+x_2^2\leq R^2$, $x_1'^2+x_2'^2\leq R^2$ and $(x_1-x'_1)^2+(x_2-x'_2)^2\leq\delta^2$ then $|f(x_1,x_2)-f(x'_1,x'_2)|\leq\varepsilon$. Then
\begin{align*}
|I(x,y,z)-f(x,y)|&\leq\frac 1{2\pi}\iint_{\mathbb R^2}\frac z{(t_1^2+t_2^2+z^2)^{3/2}}|f(t_1+x,t_2+y)-f(x,y)|dt_1dt_2\\
&\leq 2\sup_{(s_1,s_2)\in\mathbb R^2}|f(s_1,s_2)|\varepsilon+\frac 1{2\pi}\iint_{t_1^2+t_2^2\leq R^2}\frac{|z||f(t_1+x,t_2+y)-f(x,y)|}{(t_1^2+t_2^2+z^2)^{3/2}}dt_1dt_2\\
&=2\sup_{(s_1,s_2)\in\mathbb R^2}|f(s_1,s_2)|\varepsilon+ \frac 1{2\pi}\iint_{t_1^2+t_2^2\leq \delta^2}\frac{|z||f(t_1+x,t_2+y)-f(x,y)|}{(t_1^2+t_2^2+z^2)^{3/2}}dt_1dt_2\\
&+\frac 1{2\pi}\iint_{\delta^2\le t_1^2+t_2^2\leq R^2}\frac{|z||f(t_1+x,t_2+y)-f(x,y)|}{(t_1^2+t_2^2+z^2)^{3/2}}dt_1dt_2\\
&\leq\varepsilon(2\sup_{(s_1,s_2)\in\mathbb R^2}|f(s_1,s_2)|+|z|)\\
&+|z|\sup_{(s_1,s_2)\in\mathbb R^2}|f(s_1,s_2)|\iint_{R^2\geq t_1^2+t_2^2\geq \delta}\frac{dt_1dt_2}{(\delta^2+z^2)^{3/2}},
\end{align*}
so, for all $\varepsilon>0$
$$\limsup_{z\to 0}|I(x,y,z)-f(x,y)|\leq 2\varepsilon \sup_{(s_1,s_2)\in\mathbb R^2}|f(s_1,s_2)|,$$
which gives the result.