$\newcommand{\De}{\Delta}\newcommand{\vpi}{\varphi}$Let
\begin{equation*}
A:=A(y):=A(0,t,y),\quad B:=B(y):=B(0,t,y),
\end{equation*}
so that
\begin{equation*}
f(0,0,t,y)=\frac1{\sqrt{2\pi}}\,\vpi_{A(y)}(y), \quad g(0,0,t,y)=\frac1{\sqrt{2\pi}}\,\vpi_{B(y)}(y),
\end{equation*}
where
\begin{equation*}
\vpi_a(u):=\frac1{\sqrt a}\,\exp\Big(-\frac{u^2}{2a}\Big).
\end{equation*}
Note that
\begin{equation*}
\frac{\partial}{\partial a}\vpi_a(u)
=\frac12\Big(\frac{u^2}{a^{5/2}}-\frac1{a^{3/2}}\Big)\exp\Big(-\frac{u^2}{2a}\Big).
\end{equation*}
So,
\begin{equation*}
2\sqrt{2\pi}\,\int_0^\infty dy\,[f(0,0,t,y)-g(0,0,t,y)] \tag{1}\label{1}
=\int_0^1 dv\,I(v),
\end{equation*}
where
\begin{equation*}
I(v):=I_1(v)-I_2(v), \tag{2}\label{2}
\end{equation*}
\begin{equation*}
I_1(v):=\int_0^\infty dy\,H(y)
\frac{y^2}{c_v(y)^{5/2}}\exp\Big(-\frac{y^2}{2c_v(y)}\Big),
\end{equation*}
\begin{equation*}
I_2(v):=\int_0^\infty dy\,H(y)
\frac1{c_v(y)^{3/2}}\exp\Big(-\frac{y^2}{2c_v(y)}\Big),
\end{equation*}
\begin{equation*}
H:=B-A,
\end{equation*}
\begin{equation*}
c_v:=A+v(B-A).
\end{equation*}
Next is the crucial step:
\begin{equation*}
I_1(v)=I_{11}(v)+I_{12}(v), \tag{3}\label{3}
\end{equation*}
where
\begin{equation*}
I_{11}(v):=\int_0^\infty dy\,\exp\Big(-\frac{y^2}{2c_v(y)}\Big)
\Big(\frac{y}{c(y)}-\frac{y^2c'(y)}{c(y)^2}\Big)
\frac{y}{c(y)^{3/2}}H(y),
\end{equation*}
\begin{equation*}
I_{12}(v):=\int_0^\infty dy\,\exp\Big(-\frac{y^2}{2c_v(y)}\Big)
\frac{y^3c'(y)}{c(y)^{7/2}} H(y),
\end{equation*}
\begin{equation*}
c(y):=c_v(y),
\end{equation*}
and $c'$ is the derivative of the Lipschitz function $c=c_v$; this derivative exists almost everywhere (a.e.), since $k$ is $1$-Lipschitz. Moreover,
\begin{equation*}
|c'|\le t
\end{equation*}
a.e.
Note also that for $y\ge0$ and $t>0$
\begin{equation*}
t/2\le c(y)\le2t,\quad |H(y)|\ll t\,\De a,\quad |H'(y)|\ll t\,\De a,
\end{equation*}
where
\begin{equation*}
\De a:=\|a-b\|_t
\end{equation*}
and $E\ll F$ means that $|E|\le CF$ for some universal real constant $C$.
So,
\begin{equation*}
|I_{12}(v)|\ll\int_0^\infty dy\,\exp\Big(-\frac{y^2}{4t}\Big)
\frac{y^3\,t}{t^{7/2}}\, t\,\De a \asymp\sqrt t\,\De a. \tag{4}\label{4}
\end{equation*}
Integrating by parts, we have
\begin{equation*}
\begin{aligned}
I_{11}(v)&=-\int_0^\infty dy\,\Big[\frac d{dy}\exp\Big(-\frac{y^2}{2c_v(y)}\Big)\Big]
\frac{y}{c(y)^{3/2}}H(y) \\
&=\int_0^\infty dy\,\exp\Big(-\frac{y^2}{2c_v(y)}\Big)
\frac d{dy}\Big[\frac{y}{c(y)^{3/2}}H(y)\Big] \\
&=I_2(v)+I_{111}(v)-\frac32\,I_{112}(v),
\end{aligned}
\tag{5}\label{5}
\end{equation*}
where
\begin{equation*}
\begin{aligned}
I_{111}(v)&:=\int_0^\infty dy\,\exp\Big(-\frac{y^2}{2c_v(y)}\Big)
\frac{y}{c(y)^{3/2}}H'(y),
\end{aligned}
\end{equation*}
\begin{equation*}
\begin{aligned}
I_{112}(v)&:=\int_0^\infty dy\,\exp\Big(-\frac{y^2}{2c_v(y)}\Big)
\frac{yc'(y)}{c(y)^{5/2}}H(y).
\end{aligned}
\end{equation*}
Similarly to \eqref{4}, we get
\begin{equation}
|I_{111}(v)|+|I_{112}(v)|\ll \sqrt t\,\De a. \tag{6}\label{6}
\end{equation}
Collecting \eqref{1}, \eqref{2}, \eqref{3}, \eqref{4}, \eqref{5}, and \eqref{6}, we conclude that
\begin{equation*}
\Big|\int_0^\infty dy\,[f(0,0,t,y)-g(0,0,t,y)]\Big|\ll \sqrt t\,\De a,
\end{equation*}
which proves the first inequality of the two ones in question.
The other inequality can apparently be proved similarly. Since this answer is already rather long and complicated, I will leave the other inequality as an exercise (or to be posted elsewhere).
