The standard proof that I am aware of is actually explicit in this regard.

As you note, assume that $I_1\subsetneq I_2\subsetneq\cdots$ is an infinite ascending chain of ideals of $R$, let $E(R/I_n)$ be the injective envelope of $R/I_n$ for each $n$, and let $$E=\bigoplus_{n=1}^{\infty}E(R/I_n)$$
be the their direct sum.

Let $I=\bigcup\limits_{i=1}^{\infty} I_n$.

For each $n$, let $f_n$ be the composition of the embedding $I\hookrightarrow R$ with the canonical map $R\to E(R/I_n)$ (map to the quotient, then embed into the envelope).

Since we have a map from $I$ to each $E(R/I_n)$, we obtain a map $f\colon I\to \mathop{\prod}\limits_{n=1}^{\infty}E(R/I_n)$ by the universal property of the product. In fact, the image of $f$ lies in the direct sum, since for every $x\in I$ there exists $n\in\mathbb{N}$ such that $x\in I_m$ for all $m\geq n$, hence the image of $x$ is $0$ in $R/I_n$. So we have a map $I\to E$.

I claim that $f$ does not extend to a module homomorphism $\overline{f}\colon R\to E$ (which it would necessarily do if $E$ were injective). Assume to the contrary that we have an extension $\overline{f}\colon R\to E$. Being a module homomorphism with domain the free module of rank $1$, it is completely determined by $\overline{f}(1)$, and so it has the form $\overline{f}(x) = xe$ for all $x\in R$, where $e=\overline{f}(1)\in E$.

Now, let $n_0$ be a positive integer such that the $m$th component of $e$ is $0$ for all $m\geq n_0$. Let $x\in I_{n_0}\setminus I_{n_0-1}$. When we map $x$ to $R/I_{n_0}$, we obtain a nonzero element; hence the $n_0$th component of $f(x)$ must be nonzero (since $R/I_{n_0})$ embeds into $E(R/I_{n_0})$). But $f(x) = \overline{f}(x) = xe$, and the $n_0$th component of $e$ is $0$, hence so is that of $f(x)$, a contradiction.

The contradiction arises from the assumption that we can extend the map $f\colon I\to E$ to a module homomorphism $R\to E$. Hence no such extension exists, so $E$ is not injective.