I am working on solving the following pde on $[0,T]$;
$$f_t(t,x) + 2tf_x(t,x) + \frac{t^4}{2}f_{xx}(t,x) = 0 \qquad f(T,x) = x^2 = h(x)$$
By Feynman-Kac, the solution is given by
$$f(t, X_t) = E(h(X_T) \mid \mathcal{F}_t)$$
with $X_t$ a stochastic process satisfying the SDE
$$dX_t = 2t dt + t^2 dW_t$$
and hence we have
$$f(t, X_t) = E(X^2_T \mid \mathcal{F}_t)$$
How do I begin to actually compute $f(t, X_t)$? I could write
$$X_T = X_0 + \int_0^T 2s \, \mathrm{d}s + \int_0^T s^2 \, \mathrm{d}W_s$$
and write out $X^2_T$ and take the conditional expectation of that term but this gives alot of messy terms. I am probably missing something simpler, but I am complete lost to what that something might be.
It seems like you need some further clarification. What you want to do is to fix a point $x$ and a time $t$, and consider them as known. Now your solution is $$ f(t,x) = E[ h(X_T) | x,t]$$ where $X_u$ is a stochastic process satisfying \begin{equation} \begin{aligned} &\mathrm{d} X_u = 2u \, \mathrm{d}u + u^2\, \mathrm{d}W_u \quad \text{when} \quad t \leq u \leq T \\ &X_t = x \end{aligned} \end{equation}
This gives (similar to your solution, Note $ \int^T_t 2s \, \mathrm{d} s = T^2-t^2.$) $$ X_T = x + T^2-t^2 + \int^T_t s^2 \, \mathrm{d}W_s $$
Expanding $X_T^2$ is a good start.
$$ X_T^2 = (x + T^2-t^2)^2 + 2(x+T^2-t^2) \int^T_t s^2 \,\mathrm{d}W_s + \Big ( \int^T_t s^2 \, \mathrm{d}W_s \Big)^2 $$
Using the Ito isometry and that $ E[\int^T_t s^2 \, \mathrm{d}W_s] = 0$ we get $$E[X_T | \mathcal{F}_t] = (x + T^2-t^2)^2 + \int^T_t s^4\, \mathrm{d}s = (x + T^2-t^2)^2 + \frac{T^5-t^5}{5} $$