Proof.
Proof of (1). Consider the commutative diagram
\[ \xymatrix{ X' \times _ R Y \ar[rrd]^{\text{pr}'_2} \ar[rd]_{f \times \text{id}_ Y} \ar[dd]_{\text{pr}'_1} \\ & X \times _ R Y \ar[r]_{\text{pr}_2} \ar[d]_{\text{pr}_1} & Y \\ X' \ar[r]^ f & X } \]
Let $\mathcal{F}'$ be a quasi-coherent module on $X'$. We have
\begin{align*} \text{pr}_{2, *}(\text{pr}_1^*f_*\mathcal{F}' \otimes _{\mathcal{O}_{X \times _ R Y}} \mathcal{K}) & = \text{pr}_{2, *}((f \times \text{id}_ Y)_* (\text{pr}'_1)^*\mathcal{F}' \otimes _{\mathcal{O}_{X \times _ R Y}} \mathcal{K}) \\ & = \text{pr}_{2, *}(f \times \text{id}_ Y)_* \left((\text{pr}'_1)^*\mathcal{F}' \otimes _{\mathcal{O}_{X' \times _ R Y}} (f \times \text{id}_ Y)^*\mathcal{K})\right) \\ & = \text{pr}'_{2, *}((\text{pr}'_1)^*\mathcal{F}' \otimes _{\mathcal{O}_{X' \times _ R Y}} (f \times \text{id}_ Y)^*\mathcal{K}) \end{align*}
Here the first equality is affine base change for the left hand square in the diagram, see Cohomology of Schemes, Lemma 30.5.1. The second equality hold by Remark 56.5.3. The third equality is functoriality of pushforwards for modules. This proves (1).
Proof of (2). Consider the commutative diagram
\[ \xymatrix{ X \times _ R Y' \ar[rr]_-{\text{pr}'_2} \ar[rd]^{\text{id}_ X \times g} \ar[rdd]_{\text{pr}'_1} & & Y' \ar[d]^ g \\ & X \times _ R Y \ar[r]_-{\text{pr}_2} \ar[d]^{\text{pr}_1} & Y \\ & X } \]
We have
\begin{align*} g^*\text{pr}_{2, *}(\text{pr}_1^*\mathcal{F} \otimes _{\mathcal{O}_{X \times _ R Y}} \mathcal{K}) & = \text{pr}'_{2, *}( (\text{id}_ X \times g)^*( \text{pr}_1^*\mathcal{F} \otimes _{\mathcal{O}_{X \times _ R Y}} \mathcal{K})) \\ & = \text{pr}'_{2, *}((\text{pr}'_1)^*\mathcal{F} \otimes _{\mathcal{O}_{X \times _ R Y'}} (\text{id}_ X \times g)^*\mathcal{K}) \end{align*}
The first equality by flat base change for the square in the diagram, see Cohomology of Schemes, Lemma 30.5.2. The second equality by functoriality of pullback and the fact that a pullback of tensor products it the tensor product of the pullbacks.
Part (3) is a special case of (2).
$\square$
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