How to express $\partial_t^2 f(x,y)$ in partial derivatives $f_1, f_2, f_{11}, f_{12}, f_{22}$ when $x = s-2t$ and $y = s^2+t^2$? Suppose I have some random function $f(x,y)$ and I wish to represent the function $\partial_t^2 f(x,y)$ in terms of $f_1, f_2, f_{11}, f_{12}$ and $f_{22}$, when $x = s-2t$ and $y = s^2+t^2$.
My idea to how to solve this is as follows: 

Let's just work through this step by step and first note how $\nabla f = (f_1, f_2)$, and $\partial_t \vec{x} = (-2, 2t)$. Then $$\partial_t f = \nabla f \cdot \partial_t\vec{x} = -2f_1 + 2tf_2.$$ Given this we can calculate $\nabla\partial_t f = (-2f_{11}+2tf_{21},\; -2f_{12} + 2tf_{22})$. It then follows that
  $$\partial_t^2 f = \nabla \partial_t f \cdot \partial_t\vec{x} =4(f_{11} -2tf_{12}+t^2f_{22}).$$

But since the question I got this from specifically asked for all coefficients, including $f_1, f_2$, I'm not entirely sure whether this is a valid solution. If not: what am I doing wrong?
 A: I believe an issue may occur when taking $\nabla \partial_t f$. Remember that $t$ is not independent of $x$, so the $\nabla$ will not "ignore" t. (I think it is most correct to view $(x,y)$ and $(s,t)$ as two separate "coordinates")
Instead, try this. You are right that $\partial_t f(x(t)) = Df\cdot x_t$. This is a vector product. Now when we take $\partial_t^2f = \partial_t \left(Df \cdot \vec{x}_t\right)$, we use the product rule and find $\partial_t Df \cdot \vec{x}_t + Df \cdot \vec{x}_{tt}$.
Finally, we need to rewrite the $\partial_t$ with respect to $\partial_x$ and $\partial_y$. Let $g = Df$. Then $g_t = g_x x_t + g_y y_t$ by the chain rule. Now we have
$$
 x_t \partial_x(Df) \cdot \vec{x}_t + y_t \partial_y(Df) + Df \cdot \vec{x}_{tt} = 
$$
$$
-2 (f_{11},f_{21}) \cdot (-2,2t) + 2t (f_{12}, f_{22})\cdot(-2,2t)+(f_1, f_2)\cdot(0,2)
$$
I would be remiss if I didn't mention the presence of the Hessian matrix hidden here: We can factor out $\vec{x}_t$ from two terms (actually, factor it out twice), and write the Hessian matrix as $D^2f$ to see
$$
\vec{x}_t^T D^2f\vec{x}_t + Df \vec{x}_{tt}
$$
If you are brave enough with product and chain rules, maybe you can eventually see this immediately, but alas--not me on this particular date.
Best
Edit: I finally see the quick way to do it. If we avoid dot product notation and use matrix notation instead, we quickly see $\partial_t f = Df\vec{x}_t$. This by the chain rule. Then we use the product rule (and the chain rule, for the first term) to see $\partial_t^2 f = \left(\vec{x}^T_t D^2f\right)\vec{x}_t + Df \vec{x}_{tt}$ 
I don't have a great explanation for why you should put the $\vec{x}_t$ on the left when taking the second derivative. It's the only way to make the dimensions mesh, but that's a cop out.
A: You're very close. In fact, I worked it out and made the same mistake at first. If you'd like to work it out yourself: you forgot to use product rule at a certain point.
The first derivative:
$$\partial_t f(s-2t,s^2+t^2)=-2f_1(s-2t,s^2+t^2)+2tf_2(s-2t,s^2+t^2)$$
Notice the $t$ multiplying $f_2$. The second derivative:
$$4f_{11}-4tf_{12}+2f_2-4tf_{21}+4t^2f_{22}$$
$$=2(2f_{11}-4tf_{12}+f_2+2t^2f_{22})$$
