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$\newcommand\Rr{\mathbb{R}}$I am trying to show the following statement. It feels true to me, but I haven't found any reference in the literature so far:

Let $\Rr^n$ be ordered component-wise, i.e., $x \le y$ iff $x_i \le y_i$ for $i = 1,\,\ldots\,,n$.

Let f: ${\Rr}^n \to \Rr$ be a convex and increasing function. Let $h \in {\Rr^+}^n$. If $x \le x'$, then $$ f(x + h) - f(x) \;\le\; f(x' + h) - f(x'). $$

In the one-dimensional case, this is easy to see and follows e.g. from this answer. In the multi-dimensional case however, the same argument does not work any more because it might be that none of $x$, $x+h$, $x'$, $x'+h$ is a convex combination of the others. The minimal interesting case seems to be $n = 2$, $h = e_1$, and $x' = x + e_2$.

The statement is wrong if $f$ is not increasing.

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This is not true. Let $f(x_1,x_2) = \max(x_1,x_2)$. This is a convex and increasing function. Let $h = (1,0)$, $x=(0,0)$, and $x' = (1,3)$. Then $$ f(x+h)-f(x) = 1 - 0 = 1 $$ while $$ f(x'+h)-f(x') = 3 - 3 = 0 $$