For a twice differentibale mapping $f:V \to W$ between Banach spaces, if $a\in V$, then by $df_a$ I shall mean the differential of $f$ at $a$ (Frechet derivative), which is a linear transformation from $V$ into $W$. Then $df_a(h)$ shall mean the value of the linear map $df_a$ on the vector $h \in V$. In general, the chain rule says the following: if $g:U \to V$ is another twice differentiable mapping between (open subsets of) Banach spaces, then the chain rule says that
\begin{align}
d(f \circ g)_a = df_{g(a)} \circ dg_a
\end{align}
Notice that composition $(\circ)$ on the RHS is a bounded bilinear map from $L(V,W) \times L(U,V) \to L(U,W)$, defined by $(T,S) \mapsto T \circ S$, so we can apply a "generalised product rule": for all $h \in U$,
\begin{align}
d^2(f \circ g)_a[h] = \left(d^2f_{g(a)}[dg_a(h)] \right) \circ dg_a + df_{g(a)} \circ d^2g_a(h)
\end{align}
(The above is an equality of elements in $L(U,W)$). If we now use an isomorphism (which I suppress in the notation) to think of second derivatives as bilinear maps, then fully evaluating everything, we get that for all $h,k \in U$,
\begin{align}
d^2(f \circ g)_a[h,k] = d^2f_{g(a)}[dg_a(h), dg_a(k)] + df_{g(a)}[d^2g_a(h,k)]
\end{align}
(This is now an equality of elements in $W$).
I wrote the above formula of the chain rule just so you know how things work. In your specific case, they're slightly easier. For convenience, I'll only do one term; the rest are similar. Let $\phi: \Bbb{R} \to \Bbb{R}$ be defined by $\phi(t) = f(x+t,y+t)$. Then, by the chain rule,
\begin{align}
\phi'(t) = df_{(x+t,y+t)}[(1,1)]
\end{align}
Now, we can either use the second derivative formula above (in which case the second term involving $d^2g_a$ will be zero because $(1,1)$ is just a constant vector) or we can argue directly as follows. Define the "evaluation at $(1,1)$ map" $\varepsilon: L(\Bbb{R}^2, \Bbb{R}) \to \Bbb{R}$ by $\varepsilon(T) = T[(1,1)]$. Then, $\varepsilon$ itself is a bounded linear map, so for any $T$, we have
\begin{align}
d \varepsilon_T = \varepsilon \tag{$*$}
\end{align}
Using this, we can say
\begin{align}
\phi'(t) &= \varepsilon (df_{(x+t,y+t)}) \\
&= (\varepsilon \circ df)(x+t,y+t)
\end{align}
Hence, by another application of the chain rule, we find that
\begin{align}
\phi''(t) &= \left [d \varepsilon_{df_{(x+t,y+t)}} \circ d(df)_{(x+t,y+t)} \right] (1,1) \\
&= [\varepsilon \circ d^2f_{(x+t,y+t)}](1,1) \tag{by $(*)$}\\
&:= \left[d^2f_{(x+t,y+t)}(1,1) \right](1,1)
\end{align}
Now, if we make use of the isomorphism, to think of second differetials as a bilinear maps, we get that
\begin{align}
\phi''(t) = d^2f_{(x+t,y+t)} \left[ (1,1), (1,1)\right]
\end{align}
I hope you can now use this logic to compute $F''(t)$.
Additional Remarks about Notation:
By the way, just to clarify the notation, if $f:V \to W$ if differentiable at $a$, then I use $df_a$ to mean the Frechet derivative. But in the special case where the domain $V$ is just $\Bbb{R}$, then $f$ is a mapping $\Bbb{R} \to W$, so the first differential is an element $df_a \in L(\Bbb{R},W)$. But note that $L(\Bbb{R},W)$ is naturally isomorphic to $W$ via the isomorphism $T \mapsto T(1)$. So, rather than writing $df_a(1)$, I rather use the notation $f'(a)$. This is no misuse of notation, because one can show that
\begin{align}
df_a(1) = \lim_{h \to 0} \dfrac{f(a+h) - f(a)}{h}
\end{align}
Hence the use of $f'(a)$ conforms with the single variable usage.
