I'm going to switch a couple of signs and focus on $g_n(x)=\frac{1}{f_n(-x)}$ for a moment. It is an exponential generating function for the number of properly $n$-colored rooted monotonic planar forests of depth at most $1$.
I should clarify that by "$n$-colored" I mean that each edge is assigned one of given $n$ colors, and by "properly $n$-colored" I mean $n$-colored with the property that or any pair $i,j$, any edge with color $i$ is adjacent to some edge of color $j$. A monotonic rooted tree is a labelled tree for which the label of a vertex is greater than the labels of its children. It is now clear that $g_n(x)$ agrees with $e^x$ up to order $n$, since the only $g$-structures of order up to $n$ are graphs with no edges (if there were any edges, "properly $n$ colored" needs at least $n+1$ vertices).
It remains to see why the generating function of these structures is what it is. Combinatorial species give a nice framework for interpreting combinatorially what different operations on the generating function mean at the level of objects being counted. Here is a sketch:
Exponential Generating function for labelled monotonic planar rooted trees of depth at most 1 is $\log(1-x)$.
Therefore the EGF for $k$-colored labelled monotonic planar rooted trees of depth at most 1 is $\frac{1}{k}\log(1-kx)$.
The binomial tranform of order $n$ changes "$k$-colored" to "properly $n$-colored".
The exponential formula changes "tree" to "forest".
So this all gives
$$g_n(x)=\exp\left(\sum_{k=1}^n (-1)^k\binom{n}{k}\frac{1}{k}\log(1-kx)\right)$$
Here is a solution to your warm up problem. It uses a few known elementary identities, and some short inductions for identities I didn't recognize. I also changed your notation slightly from $l$ to $k$ in the internal summations.
Setting $x=2$, the first term vanishes, and we combine the second and third terms as $(B)$ and take the last term as $(A)$.
\begin{align}
&\sum_{j=0}^i \binom{n-1-j}{i-j} 2^{j-1} \sum_{k=0}^j (-1)^k \binom{n+1}{k} 2^{i-k-1}(i-k) \tag{A} \\
&\sum_{j=0}^{i-2} \binom{n-1-j}{i-2-j} 2^{j-1} \sum_{k=0}^j (-1)^k \binom{n+1}{k} \left( 2^{i-k} + 2^{i-k-1}(i-2-k)\right) \tag{B}
\end{align}
Fortunately for us, both are equal to $i\cdot2^{i-2}$ when $i$ is even, and do not depend on $n$.
First, we focus on $(A)$. Switch the order of summation to get
\begin{equation}
\sum_{k=0}^i (-1)^k \binom{n+1}{k} 2^{i-2}(i-k) \sum_{j=k}^i \binom{n-1-j}{i-j} 2^{j-k} \tag{A}
\end{equation}
Then reindex the internal summation, and repeatedly apply the hockey stick identity to show
\begin{equation}
\sum_{j=0}^{i-k}\binom{n-1-k-j}{i-k-j}2^j = \sum_{j=0}^{i-k}\binom{n-k}{j}
\end{equation}
I found it easier to write out by changing variables $a=i-k-j$, $b=i-k$ and $c=n-i-1$, then simplifying $\sum_{a=0}^b \binom{c+a}{c}2^{b-a}$.
$(A)$ naturally splits at the point $i-k$, and since we want to show that the whole thing is $i\cdot2^{i-2}$, we can break it into two pieces and show
\begin{align}
\sum_{k=1}^i (-1)^k k\binom{n+1}{k}\sum_{j=0}^{i-k} \binom{n-k}{j} &=0 \tag{A1}\\
\sum_{k=0}^i (-1)^k \binom{n+1}{k} \sum_{j=0}^{i-k} \binom{n-k}{j} &=1 \tag{A2}
\end{align}
We can simplify $(A1)$ slightly by incorporating $k$ into the binomial, and ignoring the $n+1$ term that comes out.
Switching the order of summations again, we write
\begin{equation}
Q(n,i) = \sum_{j=0}^{i-1} \sum_{k=1}^{i-j} (-1)^k \binom{n}{k-1} \binom{n-k}{j}
\end{equation}
We will induct on $n$ and $i$ to show that $Q(n,i)=0$ for $i$ even, and $-1$ for $i$ odd. The base cases are easy to check, especially if taking $i=0$ and $i=1$.
But first, we need an auxillary identity, which we will also use in $(A2)$.
\begin{equation}
P(n,i) = \sum_{j=0}^{i} (-1)^{i-j}\binom{n}{j} \binom{n-1-j}{i-j}
\end{equation}
We claim that $P(n,i)=1$ for all $n$ and $i$. This is certainly true for $i=0$. We split $\binom{n}{j}$ into two to get an induction:
\begin{align}
P(n,i) &= \sum_{j=0}^{i} (-1)^{i-j}\binom{n-1}{j} \binom{n-1-j}{i-j} + \sum_{j=1}^{i} (-1)^{i-j}\binom{n-1}{j-1} \binom{n-1-j}{i-j} \\
&= \sum_{j=0}^{i} (-1)^{i-j}\binom{n-1}{i} \binom{i}{j} + \sum_{j=1}^{i} (-1)^{i-j}\binom{n-1}{j-1} \binom{n-1-j}{i-j} \\
&= \binom{n-1}{i}\sum_{j=0}^{i} (-1)^{i-j}\binom{i}{j} + \sum_{j=0}^{i-1} (-1)^{i-1-j}\binom{n-1}{j} \binom{(n-1)-1-j}{(i-1)-j} \\
&= 0 + P(n-1,i-1)
\end{align}
Now we give a similar proof for $Q$. In the fourth to fifth lines, we used the alternating sum of binomial coefficients up to some number.
You might be able to give a direct proof if your generatingfunctionology is strong, since these look like convolutions of simple functions, but these proofs seemed easy enough that I didn't bother trying.
\begin{align}
Q(n,i) &= \sum_{j=0}^{i-1} \sum_{k=1}^{i-j} (-1)^k \binom{n}{k-1} \binom{n-k}{j} \\
&= \sum_{j=0}^{i-1} \sum_{k=1}^{i-j} (-1)^k \binom{n-1}{k-1} \binom{n-k}{j} + \sum_{j=0}^{i-2} \sum_{k=2}^{i-j} (-1)^k \binom{n-1}{k-2} \binom{n-k}{j} \\
&= \sum_{j=0}^{i-1} \sum_{k=1}^{i-j} (-1)^k \binom{n-1}{j} \binom{n-1-j}{k-1} + \sum_{j=0}^{i-2} \sum_{k=2}^{i-j} (-1)^k \binom{n-1}{k-2} \binom{n-k}{j} \\
&= \sum_{j=0}^{i-1} \binom{n-1}{j} \sum_{k=0}^{i-1-j} (-1)^{k+1} \binom{n-1-j}{k} + \sum_{j=0}^{i-2} \sum_{k=1}^{i-1-j} (-1)^{k+1} \binom{n-1}{k-1} \binom{n-1-k}{j} \\
&= \sum_{j=0}^{i-1} \binom{n-1}{j} (-1)^{i-1-j+1} \binom{n-2-j}{i-1-j} - Q(n-1, i-1) \\
&= -P(n-1,i-1) - Q(n-1, i-1)
\end{align}
Now, since our equation $(A1)$ was just $(n+1)Q(n,i)$, and $i$ is even, it's $0$. A similar treatment yields $(A2)$, again using the alternating binomial sum identity near the end:
\begin{align}
&\,\sum_{k=0}^i (-1)^k \binom{n+1}{k} \sum_{j=0}^{i-k} \binom{n-k}{j} \\
&= \sum_{j=0}^i \sum_{k=0}^{i-j} (-1)^k \binom{n+1}{k} \binom{n-k}{j} \\
&= \sum_{j=0}^i \sum_{k=0}^{i-j} (-1)^k \binom{n}{k} \binom{n-k}{j} + \sum_{j=0}^i \sum_{k=0}^{i-j} (-1)^k \binom{n}{k-1} \binom{n-k}{j} \\
&= \sum_{j=0}^i \sum_{k=0}^{i-j} (-1)^k \binom{n}{j} \binom{n-j}{k} + Q(n,i) \\
&= \sum_{j=0}^i \binom{n}{j} \sum_{k=0}^{i-j} (-1)^k \binom{n-j}{k} \\
&= \sum_{j=0}^i \binom{n}{j} (-1)^{i-j} \binom{n-1-j}{i-j} \\
&= P(n,i)
\end{align}
So we have shown that $(A) = i\cdot2^{i-2}$ for $i$ even. Let's use this for $(B)$, since $i-2$ is also even, we have:
\begin{equation}
\sum_{j=0}^{i-2} \binom{n-1-j}{i-2-j} 2^{j-1} \sum_{k=0}^j (-1)^k \binom{n+1}{k} 2^{i-k-3}(i-2-k) = (i-2)2^{i-4}
\end{equation}
After multiplying both sides by $2^2$, the only place the left side differs from $(B)$ is the extra $-2$ in $(i-2-k)$, and the right side is $2^{i-1}$ smaller than we would like.
So we show that
\begin{equation}
\sum_{j=0}^{i-2} \binom{n-1-j}{i-2-j} 2^{j-1} \sum_{k=0}^j (-1)^k \binom{n+1}{k} 2^{i-k} = 2^{i-1}
\end{equation}
Dividing out the factor of $2^{i-1}$ and switching the order of summation, we get
\begin{equation}
\sum_{k=0}^{i-2} (-1)^k \binom{n+1}{k} \sum_{j=k}^{i-2} \binom{n-1-j}{i-2-j}2^{j-k}
\end{equation}
Of course we recognize our hockey stick identity from earlier, so this simplifies to the case of $(A2)$
\begin{equation}
\sum_{k=0}^{i-2} (-1)^k \binom{n+1}{k} \sum_{j=0}^{i-2-k} \binom{n-k}{j} =1
\end{equation}
Best Answer
I don't have an answer, but I have spend a couple of hours on it, so here is some of my thoughts on the problem.
In the following I will identify expressions with sets, so $2^n$ corresponds to the set of 01-sequences of length n, $\binom{n}{k}$ is the set of 01-sequence of length n with exactly k 1s, and products and sums corresponds to taking product sets and unions.
We want to find a bijective function from $\sum_{k=0}^m 2^{2(m-k)} \binom{2k}{k} \binom{2m-k}{m}$ to $\binom{4m+1}{2m}$ (I have multiplied with $4^m$ on both sides). Let me give an example a similar looking equality (you can skip the rest of this paragraph if you want): $\sum_{k=0}^m2^{2(m-k)}\binom{2k}{k}\binom{2m}{2k}=\binom{4m}{2m}$. Take an element in $\binom{4m}{2m}$, and pair the terms, so we have a sequence over {(00),(01),(10),(11)} of length $2m$. We have an even number of 1s in the sequence, so there must be an even number of pairs that contain exactly one 1, and thus and even number of pairs with (00) or (11). Let the number of (00) and (11) in the sequence of pairs be 2k. Now k of these must be (00) and k of them is (11) is the number of 0s and 1s are the same. The 2k terms in the 2m length sequence can be chosen in $\binom{2m}{2k}$ ways, the k (11)s of these 2k terms can chosen in $\binom{2k}{k}$ ways and in the rest of the $2m-2k$ terms we must choose between (10) and (01). This gives a factor $2^{2(m-k)}$, so we have a 1-1 correspondence between $\sum_{k=0}^m2^{2(m-k)}\binom{2k}{k}\binom{2m}{2k}$ and $\binom{4m}{2m}$.
An important part of such a proof is to find out what k represents. In your equality, it turns out that the k=0 part of the sum is about $\sqrt{\frac{1}{2}}$ of the whole sum. Do anyone know where the $\sqrt{\frac{1}{2}}$ could come from? One way to find out what k is, would be to find an injective function from the k=0 term, $2^{2m}\binom{2m}{m}$, to $\binom{4m+1}{2m}$ and see what the image set looks like. But I haven't been able to find such a function, that is, I cannot find a combinatorial proof that $2^{2m}\binom{2m}{m}\leq \binom{4m+1}{2m}$ (nor that $\binom{4m}{2m}<2^{2m}\binom{2m}{m}$). Perhaps you should try to ask this in you question?