Proof.
Before giving the proofs, we explain the meaning of (c). Suppose we have an additional cartesian square
\[ \xymatrix{ X'' \ar[d]_{f''} \ar[r]_{h'} & X' \ar[d]_{f'} \ar[r]_{g'} & X \ar[d]^ f \\ Y'' \ar[r]^ h & Y' \ar[r]^ g & Y } \]
tacked onto our given diagram. If (a) holds, then there is a canonical map $\gamma : h^*R^ df'_*\mathcal{F}' \to R^ df''_*(h')^*\mathcal{F}'$. Namely, $\gamma $ is the map on degree $d$ cohomology sheaves induced by the composition
\[ Lh^*Rf'_*\mathcal{F}' \longrightarrow Rf''_*L(h')^*\mathcal{F}' \longrightarrow Rf''_*(h')^*\mathcal{F}' \]
Here the first arrow is the base change map (Cohomology, Remark 20.28.3) and the second arrow complex from the canonical map $L(g')^*\mathcal{F} \to (g')^*\mathcal{F}$. Similarly, since $Rf'_*\mathcal{F}$ has no nonzero cohomology sheaves in degrees $> d$ by (a) we have $H^ d(Lh^*Rf_*\mathcal{F}') = h^*R^ df_*\mathcal{F}$. The content of (c) is that $\gamma $ is an isomorphism.
Having said this, we can check (a), (b), and (c) locally on $Y'$ and $Y''$. Suppose that $V \subset Y$ is a quasi-compact open subscheme. Then we claim (1) and (2) hold for $f|_{f^{-1}(V)} : f^{-1}(V) \to V$. Namely, (1) is immediate and (2) follows because any quasi-coherent module on $f^{-1}(V)$ is the restriction of a quasi-coherent module on $X$ (Properties, Lemma 28.22.1) and formation of higher direct images commutes with restriction to opens. Thus we may also work locally on $Y$. In other words, we may assume $Y''$, $Y'$, and $Y$ are affine schemes.
Proof of (a) when $Y'$ and $Y$ are affine. In this case the morphisms $g$ and $g'$ are affine. Thus $g_* = Rg_*$ and $g'_* = Rg'_*$ (Cohomology of Schemes, Lemma 30.2.3) and $g_*$ is identified with the restriction functor on modules (Schemes, Lemma 26.7.3). Then
\[ g_*(R^ if'_*\mathcal{F}') = H^ i(Rg_*Rf'_*\mathcal{F}') = H^ i(Rf_*Rg'_*\mathcal{F}') = H^ i(Rf_*g'_*\mathcal{F}') = Rf^ i_*g'_*\mathcal{F}' \]
which is zero by assumption (2). Hence (a) by our description of $g_*$.
Proof of (b) when $Y'$ is affine, say $Y' = \mathop{\mathrm{Spec}}(R')$. By part (a) we have $H^{d + 1}(X', \mathcal{F}') = 0$ for any quasi-coherent $\mathcal{O}_{X'}$-module $\mathcal{F}'$, see Cohomology of Schemes, Lemma 30.4.6. Consider the functor $F$ on $R'$-modules defined by the rule
\[ F(M) = H^ d(X', \mathcal{F}' \otimes _{\mathcal{O}_{X'}} (f')^*\widetilde{M}) \]
By Cohomology, Lemma 20.19.1 this functor commutes with direct sums (this is where we use that $X$ and hence $X'$ is quasi-compact and quasi-separated). On the other hand, if $M_1 \to M_2 \to M_3 \to 0$ is an exact sequence, then
\[ \mathcal{F}' \otimes _{\mathcal{O}_{X'}} (f')^*\widetilde{M}_1 \to \mathcal{F}' \otimes _{\mathcal{O}_{X'}} (f')^*\widetilde{M}_2 \to \mathcal{F}' \otimes _{\mathcal{O}_{X'}} (f')^*\widetilde{M}_3 \to 0 \]
is an exact sequence of quasi-coherent modules on $X'$ and by the vanishing of higher cohomology given above we get an exact sequence
\[ F(M_1) \to F(M_2) \to F(M_3) \to 0 \]
In other words, $F$ is right exact. Any right exact $R'$-linear functor $F : \text{Mod}_{R'} \to \text{Mod}_{R'}$ which commutes with direct sums is given by tensoring with an $R'$-module (omitted; left as exercise for the reader). Thus we obtain $F(M) = H^ d(X', \mathcal{F}') \otimes _{R'} M$. Since $R^ d(f')_*\mathcal{F}'$ and $R^ d(f')_*(\mathcal{F}' \otimes _{\mathcal{O}_{X'}} (f')^*\widetilde{M})$ are quasi-coherent (Cohomology of Schemes, Lemma 30.4.5), the fact that $F(M) = H^ d(X', \mathcal{F}') \otimes _{R'} M$ translates into the statement given in (b).
Proof of (c) when $Y'' \to Y' \to Y$ are morphisms of affine schemes. Say $Y'' = \mathop{\mathrm{Spec}}(R'')$ and $Y' = \mathop{\mathrm{Spec}}(R')$. Then we see that $R^ df''_*(h')^*\mathcal{F}'$ is the quasi-coherent module on $Y'$ associated to the $R''$-module $H^ d(X'', (h')^*\mathcal{F}')$. Now $h' : X'' \to X'$ is affine hence $H^ d(X'', (h')^*\mathcal{F}') = H^ d(X, h'_*(h')^*\mathcal{F}')$ by the already used Cohomology of Schemes, Lemma 30.2.4. We have
\[ h'_*(h')^*\mathcal{F}' = \mathcal{F}' \otimes _{\mathcal{O}_{X'}} (f')^*\widetilde{R''} \]
as the reader sees by checking on an affine open covering. Thus $H^ d(X'', (h')^*\mathcal{F}') = H^ d(X', \mathcal{F}') \otimes _{R'} R''$ by part (b) applied to $f'$ and the proof is complete.
$\square$
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