R noetherian implies that any direct limit of injective modules is injective

I am not convinced that your argument is correct. It's written out a bit confusingly in spots, which makes the proof difficult to evaluate. One would usually write about taking an upper bound of the $k_j$ and not the max, for example. I am still not clear on what the $A_i$ are. You reserve $i$ for morphisms to the direct limit and then use it as an index for your directed system. That sort of thing.

Now for the bit about the $B_i$ specifically, I don't really understand the argument about surjectivity, but I suspect the following is a counterexample. Take the category of $\mathbb Z$-modules and consider $Z := \oplus_{\mathbb N} \mathbb Z$. Let $\sigma: Z \to Z$ be the left shift homomorphism (note that this erases the leftmost coordinate). We can regard this as a directed set $Z \overset{\sigma}{\to} Z \overset{\sigma}{\to} Z \overset{\sigma}{\to} \dots$, which has direct limit zero (since enough applications of $\sigma$ will erase all of the finitely many nonzero coordinates of any element), and for which every one of the directed system maps is surjective. We even have that $\mathbb Z$ is Noetherian, so the situation is quite similar, and still the modules do not 'stabilize' at zero.

Here is an approach of the 'hands on' sort you wanted to use: take your $a_1, \dots, a_n$ and pull $f(a_1), \dots, f(a_n)$ back to elements $y_1, \dots, y_n$ in some module $E_\alpha$ of your directed system. We would like to send $a_i \mapsto y_i$ and have this extend linearly to a map $\phi: I \to E_\alpha$. Unfortunately this extension is probably not well defined. You can check that if $\sum \lambda_i a_i = 0$ implies $\sum \lambda_i y_i = 0$, then $\phi$ is well defined.

Let us say that we transport an element $x$ of some $E_i$ to $E_j$ if we apply some $h^i_j$ to $x$ and regard $h^i_j(x)$ as the 'new' $x$.

Notice that if we have the insertion $\rho_\alpha: E_\alpha \to L$ with $\rho_\alpha(x) = 0$, then we can transport $x$ to some module where it will become $0$. Now the $\vec \lambda \in R^n$ such that $\sum \lambda_i a_i = 0$ form a submodule, which since $R$ is Noetherian is finitely generated, say by $\Lambda^1, \dots, \Lambda^z$ (these are indices, not exponents). Since $$\rho_\alpha(\sum \Lambda^1_i y_i) = \sum \Lambda^1_i \rho_\alpha(y_i) = \sum \Lambda^1_i f(a_i) = f(\sum \Lambda^1_i a_i) = 0$$ it must be that we can transport the $y_i$ so that $\sum \Lambda^1_i y_i = 0$. Repeat this with $\Lambda^2, \dots, \Lambda^z$ to find a module where the desired $\phi$ is well defined. You have already demonstrated you know how to complete the rest from here.