It is possible to give a direct combinatorial proof, but it is quite difficult to find it.
One possibility is to use paths between points with integer coordinates and steps $(1,1)$ and $(1,-1)$.
1) $\binom{2n}{n}$ counts all paths from $(0,0)$ to $(2n,0)$.
2) $2^{2n}$ counts all paths starting from $(0,0)$ with $2n$ steps.
3) $\binom{2n}{n}$ counts all paths with $2n$ steps that never touch the $x$-axis again after the start. (This one is not obvious, but can be proved
with a bijection.)
Now you can conclude that all paths are a concatenation of a path that returns a certain number of times to the $x$-axis and a path that never does.
Note that the main difficulty here was that the two binomial coefficients are interpreted differently.
Edited to add reference:
In Richard P. Stanley: Enumerative Combinatorics Volume 1, Chapter 1, Solution to exercice 2c the following reference is given:
The problem of giving a combinatorial proof was raised by P. Veress and solved by G. Hajos in the 1930s. A recent proof appears in D.J. Kleitman, Studies in Applied Math. 54 (1975), 289 - 292. See also M. Sved, Math. Intelligencer, vol.6, no. 4 (1984), 44-45.
But I have not looked to check which article gives the proof I have outlined above.
Starting from (the contribution from $k=0$ is zero owing to the third
binomial coefficient)
$$\sum_{k=1}^n \left(-\frac{1}{4}\right)^k
{2k\choose k}^2 \frac{1}{1-2k} {n+k-2\choose 2k-2}$$
we seek to show that this is zero when $n\gt 1$ is odd and
$$\left[\left(\frac{1}{4}\right)^m
{2m\choose m} \frac{1}{1-2m}\right]^2$$
when $n=2m$ is even.
We observe that with $k\ge 1$
$${2k\choose k} \frac{1}{1-2k} {n+k-2\choose 2k-2}
= 2 {2k-1\choose k-1} \frac{1}{1-2k} {n+k-2\choose 2k-2}
\\ = -2 {2k-2\choose k-1} \frac{1}{k} {n+k-2\choose 2k-2}
= -\frac{2}{k} \frac{(n+k-2)!}{(k-1)!^2 \times (n-k)!}
\\ = -\frac{2}{k} {n+k-2\choose k-1} {n-1\choose k-1}
= -\frac{2}{n} {n\choose k} {n+k-2\choose k-1}.$$
We get for our sum
$$-\frac{2}{n} \sum_{k=1}^n
{n\choose k} \left(-\frac{1}{4}\right)^k
{2k\choose k}
{n+k-2\choose k-1}
\\ = -\frac{2}{n} \sum_{k=1}^n
{n\choose k} {-1/2\choose k}
{n+k-2\choose n-1}
\\ = -\frac{2}{n} [z^{n-1}] (1+z)^{n-2} \sum_{k=1}^n
{n\choose k} {-1/2\choose k} (1+z)^k.$$
The value $k=0$ contributes zero:
$$-\frac{2}{n} \times
\;\underset{w}{\mathrm{res}}\; \frac{1}{w} (1+w)^{-1/2}
[z^{n-1}] (1+z)^{n-2} \sum_{k=0}^n
{n\choose k} \frac{1}{w^k} (1+z)^k
\\ = -\frac{2}{n} \times
\;\underset{w}{\mathrm{res}}\; \frac{1}{w} (1+w)^{-1/2}
[z^{n-1}] (1+z)^{n-2} (1+(1+z)/w)^n
\\ = -\frac{2}{n} \times
\;\underset{w}{\mathrm{res}}\; \frac{1}{w^{n+1}} (1+w)^{-1/2}
[z^{n-1}] (1+z)^{n-2} (1+w+z)^n
\\ = -\frac{2}{n} \times
\;\underset{w}{\mathrm{res}}\; \frac{1}{w^{n+1}} (1+w)^{-1/2}
[z^{n-1}] (1+z)^{n-2}
\sum_{q=0}^n {n\choose q} (1+w)^q z^{n-q}
\\ = -\frac{2}{n} \times
\sum_{q=1}^n {n\choose q} {q-1/2\choose n}
{n-2\choose q-1}.$$
Now observe that with $q\lt n$ (third binomial coefficient is zero
when $q=n$)
$${q-1/2\choose n} = \frac{1}{n!} (q-1/2)^\underline{n}
= \frac{1}{n!} \prod_{p=0}^{q-1} (q-1/2-p)
\prod_{p=q}^{n-1} (q-1/2-p)
\\ = \frac{1}{n! \times 2^n} \prod_{p=0}^{q-1} (2q-1-2p)
\prod_{p=q}^{n-1} (2q-1-2p)
\\ = \frac{1}{n! \times 2^n}
\frac{(2q-1)!}{(q-1)! \times 2^{q-1}}
\prod_{p=0}^{n-1-q} (-1-2p)
\\ = \frac{(-1)^{n-q}}{n! \times 2^n}
\frac{(2q-1)!}{(q-1)! \times 2^{q-1}}
\frac{(2n-1-2q)!}{(n-1-q)! \times 2^{n-1-q}}
\\= \frac{(-1)^{n-q}}{2^{2n-2}}
{n\choose q}^{-1} {2q-1\choose q-1} {2n-1-2q\choose n-q}.$$
We get for our sum
$$-\frac{1}{n \times 2^{2n-3}} \times
\sum_{q=1}^{n-1} (-1)^{n-q}
{2q-1\choose q-1} {2n-1-2q\choose n-q}
{n-2\choose q-1}
\\ = \frac{1}{n \times 2^{2n-3}} \times
\sum_{q=0}^{n-2} {n-2\choose q} (-1)^{n-2-q}
{2q+1\choose q} {2n-3-2q\choose n-q-1}.$$
This becomes
$$\frac{1}{n \times 2^{2n-3}} \times [z^{n-1}] (1+z)^{2n-3}
\sum_{q=0}^{n-2} {n-2\choose q} (-1)^{n-2-q}
{2q+1\choose q} z^q (1+z)^{-2q}
\\ = \frac{1}{n \times 2^{2n-3}}
\;\underset{w}{\mathrm{res}}\; \frac{1+w}{w} [z^{n-1}] (1+z)^{2n-3}
\\ \times
\sum_{q=0}^{n-2} {n-2\choose q} (-1)^{n-2-q}
\frac{1}{w^{q}} (1+w)^{2q} z^q (1+z)^{-2q}
\\ = \frac{1}{n \times 2^{2n-3}}
\;\underset{w}{\mathrm{res}}\; \frac{1+w}{w} [z^{n-1}] (1+z)^{2n-3}
\left(\frac{z(1+w)^2}{w(1+z)^2}-1\right)^{n-2}
\\ = \frac{1}{n \times 2^{2n-3}}
\;\underset{w}{\mathrm{res}}\; \frac{1+w}{w^{n-1}} [z^{n-1}] (1+z)
\left(z(1+w)^2-w(1+z)^2\right)^{n-2}
\\ = \frac{1}{n \times 2^{2n-3}}
\;\underset{w}{\mathrm{res}}\; \frac{1+w}{w^{n-1}} [z^{n-1}] (1+z)
(z-w)^{n-2} (1-wz)^{n-2}.$$
The first piece in $z$ is
$$[z^{n-1}] (z-w)^{n-2} (1-wz)^{n-2}
\\ = \sum_{q=1}^{n-2} {n-2\choose q} (-1)^{n-2-q} w^{n-2-q}
{n-2\choose n-1-q} (-1)^{n-1-q} w^{n-1-q}
\\ = - \sum_{q=1}^{n-2} {n-2\choose q} {n-2\choose q-1}
w^{2n-3-2q}.$$
Here we require
$$([w^{n-2}] + [w^{n-3}]) w^{2n-3-2q}$$
We get $q=(n-1)/2$ in the first case and $q=n/2$ in the second. As
this is a pair of an integer and a fraction clearly only one of these
extractors can return a non-zero value.
The second piece in $z$ is
$$[z^{n-2}] (z-w)^{n-2} (1-wz)^{n-2}
\\ = \sum_{q=0}^{n-2} {n-2\choose q} (-1)^{n-2-q} w^{n-2-q}
{n-2\choose n-2-q} (-1)^{n-2-q} w^{n-2-q}
\\ = \sum_{q=0}^{n-2} {n-2\choose q} {n-2\choose q}
w^{2n-4-2q}.$$
Solving for $q$ again we require
$$([w^{n-2}] + [w^{n-3}]) w^{2n-4-2q}$$
getting $q=n/2-1$ and $q=(n-1)/2.$
Supposing that $n$ is odd i.e. $n=2m+1$ we thus have
$$-{2m-1\choose m} {2m-1\choose m-1} +
{2m-1\choose m} {2m-1\choose m} = 0,$$
and we have proved the second part of the claim.
On the other hand with $n=2m$ even we collect
$$-{2m-2\choose m} {2m-2\choose m-1}
+ {2m-2\choose m-1} {2m-2\choose m-1}
\\ = {2m-2\choose m-1}^2 \left(1 - \frac{m-1}{m}\right)
= \frac{m^2} {(2m-1)^2} {2m-1\choose m}^2
\frac{1}{m}
\\ = \frac{m^2} {(2m-1)^2} \frac{m^2}{(2m)^2} {2m\choose m}^2
\frac{1}{m}
= \frac{1}{4} \frac{m} {(2m-1)^2} {2m\choose m}^2.$$
Restoring the factor in front we obtain
$$\frac{1}{n \times 2^{2n-3}}
\frac{1}{4} \frac{m} {(2m-1)^2} {2m\choose m}^2
= \frac{1}{2^{2n}} \frac{1} {(2m-1)^2} {2m\choose m}^2
\\ = \frac{1}{2^{4m}} \frac{1} {(1-2m)^2} {2m\choose m}^2$$
This is
$$\bbox[5px,border:2px solid #00A000]{
\left[\left(\frac{1}{4}\right)^m
{2m\choose m} \frac{1}{1-2m}\right]^2}$$
as was to be shown.
Best Answer
Your equation (2) can be rewritten as $$\sum_{k=1}^{m-1} C_{k-1}C_{m-k-1} = C_{m-1},$$ where $C_n$ is the $n^{\text{th}}$ Catalan number. Any introductory article on Catalan numbers should contain one or more counting proofs of that identity.