The issue is with notation and abuse of language. First of all $\theta$ is a function of 3 variables, so writing $\frac{d\theta}{dt}$ is just wrong. Yes, people write it all the time, but it is wrong. The two $\theta$'s mean completely different things on the two sides of the equation.
The chain rule tells us the following. Suppose we have a (differentiable) parametrized curve $\gamma:\Bbb{R}\to\Bbb{R}^2$. Then, we can construct a NEW function $\Theta:\Bbb{R}\to\Bbb{R}$ as $\Theta(t)=\theta(t,\gamma(t))=\theta(t,\gamma_1(t),\gamma_2(t))$. It may probably be a good idea to write $\Theta_{\gamma}$ to emphasize that this function is constructed from $\theta$ and the curve $\gamma$... but I won't simply for ease of typing. Then,
\begin{align}
\Theta'(t)&=(\partial_1\theta)_{(t,\gamma_1(t),\gamma_2(t))}\cdot 1 +
(\partial_2\theta)_{(t,\gamma_1(t),\gamma_2(t))}\cdot \gamma_1'(t)+
(\partial_3\theta)_{(t,\gamma_1(t),\gamma_2(t))}\cdot \gamma_2'(t)\tag{i}
\end{align}
Or, written in slightly more classical notation,
\begin{align}
\frac{d\Theta}{dt}\bigg|_t&=\frac{\partial \theta}{\partial t}\bigg|_{(t,\gamma_1(t),\gamma_2(t))}+
\frac{\partial \theta}{\partial x}\bigg|_{(t,\gamma_1(t),\gamma_2(t))}\cdot
\frac{d\gamma_1}{dt}\bigg|_t+
\frac{\partial \theta}{\partial y}\bigg|_{(t,\gamma_1(t),\gamma_2(t))}\cdot
\frac{d\gamma_2}{dt}\bigg|_t\tag{ii}
\end{align}
Or if you go one step further and omit the points of evaluation (which are actually important, since this equation expresses what happens only along the curve $\gamma$), we can write
\begin{align}
\frac{d\Theta}{dt}&=\frac{\partial \theta}{\partial t}+
\frac{\partial \theta}{\partial x}\frac{d\gamma_1}{d t}+
\frac{\partial \theta}{\partial y}\frac{d\gamma_2}{d t}\tag{iii}
\end{align}
The sloppiest of all is
\begin{align}
\frac{d\theta}{dt}&=\frac{\partial \theta}{\partial t}+
\frac{\partial \theta}{\partial x}\frac{dx}{d t}+
\frac{\partial \theta}{\partial y}\frac{dy}{d t},\tag{iv}
\end{align}
because the $\theta$ on the LHS really should be $\Theta$ which is a completely different function, and the symbol $x$ has two meanings. One is an arbitrary placeholder for the second entry of the function $\theta$, and also as the first component of the parametrized curve. Similarly, $y$ has two meanings. ALso, the points of evaluation are not indicated at all.
If you're still not convinced of the dangers of such condensed notation/ are unfamiliar with it, perhaps Partial Derivatives of Functions of Functions might help.
Getting directly to the case you're asking about, let us fix a point $(a,b)\in\Bbb{R}^2$, and suppose we consider the constant curve $\gamma:\Bbb{R}\to\Bbb{R}^2$, defined as $\gamma(t)=(a,b)$. Now, if you plug this into equations (i) or (ii), you'll see that the function $\Theta$ is defined as $\Theta(t):=\theta(t,\gamma(t))=\theta(t,a,b)$, and that its derivative is given by
\begin{align}
\frac{d\Theta}{dt}\bigg|_t&=\frac{\partial \theta}{\partial t}\bigg|_{(t,a,b)}+0+0.
\end{align}
Well, this should be obvious even from the very definition of a partial derivative. The meaning of the partial derivative symbol $\frac{\partial \theta}{\partial t}\bigg|_{(t,a,b)}$ or $(\partial_1\theta)_{(t,a,b)}$ means we fix all the other entries, so that we're left with a single-variable function $\Theta(\cdot)=\theta(\cdot, a,b)$, and then calculate the derivative of this function at $t$. This is literally definition. There's no need to talk about fields/particle motion or anything else.
Note however, that you should not fall into the trap of saying $\frac{d\Theta}{dt}=\frac{\partial \theta}{\partial t}$. On the LHS, $\Theta:\Bbb{R}\to\Bbb{R}$ is a function of one variable, so its derivative $\Theta'=\frac{d\Theta}{dt}:\Bbb{R}\to\Bbb{R}$ is also a function of one variable. On the RHS, $\theta:\Bbb{R}^3\to\Bbb{R}$ is a function of 3 variables, hence its partial derivative $\partial_1\theta=\frac{\partial \theta}{\partial t}:\Bbb{R}^3\to\Bbb{R}$
is also a function of 3 variables. A function of one-variable can never be equal to a function of three variables.
So, you adivsor's claim that $\frac{d\theta}{dt}=\frac{\partial \theta}{\partial t}$ (or even with my slightly more precise notation, the "equation" $\frac{d\Theta}{dt}=\frac{\partial \theta}{\partial t}$) are strictly speaking wrong. They are only equal if you know what you're talking about.
