Sorry I haven't read your proof (it looks a little bit clumsy). I would prove it as follows:
Locally of finite type morphisms are closed under composition. Hence, it suffices to prove that closed immersions and open immersions are of finite type. For closed immersions, this is clear: Simply use that $A/I$ is a finitely generated $A$-algebra for every ideal $I$ of a commutative ring $A$ (in fact, $\emptyset$ is a generating system!).
Now let $U \to X$ be an open immersion. By may work locally on $X$, so let us assume that $X=\mathrm{Spec}(A)$ is affine. Then $ U = \cup_i D(f_i)$ is a union of basic-open subsets. Each $D(f_i)$ is affine and of finite type over $X$, since $A_{f_i}$ is a finitely generated $A$-algebra (namely, generated by $f_i^ {-1}$). Hence, $U \to X$ is locally of finite type.
I think that the proof of this is actually not so hard, but it is rather long and fiddly. I also think that being able to reproduce this proof is actually a pretty good indicator for whether or not you really understand the general procedure of showing that some property of morphisms holds for some affine cover iff it holds for every affine cover, since they're all more or less the same apart from the interchanging of some algebraic results at the end. Anyway, here's the proof:
First recall the very important fact that given any two affine open subschemes $U$ and $V$ of a given scheme $X$, their intersection can be covered by open subchemes that are simultaneously distinguished open affine subschemes for $U$ and for $V$. This will be used multiple times in this proof, as well as many others.
Thus we may cover $V$ by open subschemes that are simultaneously distinguished affine open subschemes for $V$ and some $V_i$. Since the collection of all such subschemes cover $V$, the collection of all such pre-images will cover $f^{-1}(V)$. Let $V'$ be one such open subscheme, say $V' \cong D(g), D(g_i)$ where $g \in B$ and $g_i \in B_i$. We may further cover $f^{-1}(V')$ by open subschemes which are each distinguished open affine for some $U_{i,j}$. Let $U'$ be one such, say $U' \cong D(f_{i,j})$ for some $f_{i,j} \in A_{i,j}$.
Now, the restriction of $f:U_{i,j}\rightarrow V_i$ is induced by a ring map $\varphi_{i,j}:B_i\rightarrow A_{i,j}$ making $A_{i,j}$ a finitely generated $B_i-$algebra. Since $U' \subset f^{-1}(V\cap V_i)$, we know that $f:U'\rightarrow V_i$ is induced by the (localization of the) map $\varphi_{i,j}$, but also that it is induced by a map $\varphi:B\rightarrow (A_{i,j})_{f_{i,j}}$. We will show that this makes $A:=(A_{i,j})_{f_{i,j}}$ into a finitely generated $B$-algebra and we will be done.
We already used the fact that $U' \subset f^{-1}(V)$ and $U' \subset f^{-1}(V_i)$ to get the ring maps $\varphi$ and $\varphi_{i,j}$, but we also have that $U' \subset f^{-1}(V')$, which gives us two more ring maps induced by $f$, $\psi:B_g \rightarrow A$ and $\psi_{i,j}:(B_i)_{g_i} \rightarrow A$. By the construction of $V'$ and $U'$, it follows that $\psi$ is the localization of $\varphi$ and $\psi_{i,j}$ is the localization of $\varphi_{i,j}$ composed with the localization map $A_{i,j} \rightarrow A$. Moreover, these two maps commute with the isomorphism $B_g \cong (B_i)_{g_i}$.*
Now, the coup de grâce: by assumption $A_{i,j}$ was a finitely generated $B_i$-algebra and so $A=(A_{i,j})_{f_{i,j}}$ is a finitely generated $B_i$-algebra. But then it is a finitely generated $(B_i)_{g_i}$-algebra, which is equivalent to being a finitely generated $B_g$-algebra. Finally, being a finitely generated $B_g$-algebra implies that it is a finitely generated $B$-algebra, which is what we were looking for, so we win.
Remark: You can perhaps streamline the proof a little by taking $U' = U_{i,j} \cap f^{-1}(V')$ but I don't think this makes the proof any easier or quicker.
*This can be tricky to see, although it may also be completely obvious. Either way, I recommend drawing a picture of all of the open affine schemes involved, showing where each maps to under $f$ and which are contained in others. You can then draw the corresponding diagram of rings and maps between them to see why this is true. It is important to actually do this step, though it is often overlooked, since $A$ being a finitely generated $B$ algebra is a property dependent on the map from $B$ to $A$, so if we don't know what the ring maps are, then you can't possibly complete the proof.
Best Answer
If $B$ is a ring and $f\in B$, then $B_f=B[x]/(xf-1)$.
In your case, this gives that $R[\underline{t}]_f=R[\underline{t},x]/(fx-1)$. Now for any ideal in this ring, i.e. an ideal $\mathfrak{a}$ of $R[\underline{t},x]$ containing $fx-1$, you obtain that its quotient is $R[\underline{t},x]/\mathfrak{a}$. And if $\mathfrak{a}/(fx-1)$ is finitely generated as an ideal of $R[\underline{t},x]/(fx-1)$, then so will be $\mathfrak{a}$ as an ideal of $R[\underline{t},x]$.