Is it possible to find a closed-form expression for this summation?
$$\sum\limits_{k = 0}^n {\left( \left( {\matrix{ 2n \cr k \cr } } \right)+ \left( {\matrix{ 2n \cr n+k-1 \cr } } \right) \left( \frac{1}{3}\right)^\left(2n-2k+1 \right) \right) \cdot\left( \frac{1}{3}\right)^k}$$
I can not find one that is not using hyper geometric functions, and splitting the sum did not help me, so there might be something I am missing. If anyone has advice that would be great.
When you have this kind of summation, you cannot avoid at least gaussian hypergeometric functions and in most cases you will not obtain any closed form.
Relacing the $\frac 13$ by $x$ and splitting the sum, you have $$\sum_{k=0}^n \binom{2 n}{k} x^k=(x+1)^{2 n}-\binom{2 n}{n+1} x^{n+1} \, _2F_1(1,1-n;n+2;-x)$$ The second term will be $$\sum_{k=0}^n \binom{2 n}{n+1-k} x^{2 n-k}=\binom{2 n}{n+1} x^{2 n} \, _2F_1\left(1,-n-1;n;-\frac{1}{x}\right)-x^{n-1}$$ probably leading to what Wolfram Alpha gave for your specific case.
Now, for a given value of $n$, what you will obtain is a polynomial of degree $2n$ in $x$ with integer coefficients. The very first ones are
$$\left( \begin{array}{cc} n & \text{Polynomial} \\ 1 & x^2+4 x+1 \\ 2 & 4 x^4+6 x^3+10 x^2+4 x+1 \\ 3 & 15 x^6+20 x^5+15 x^4+26 x^3+15 x^2+6 x+1 \\ 4 & 56 x^8+70 x^7+56 x^6+28 x^5+78 x^4+56 x^3+28 x^2+8 x+1 \\ 5 & 210 x^{10}+252 x^9+210 x^8+120 x^7+45 x^6+262 x^5+210 x^4+120 x^3+45 x^2+10 x+1 \end{array} \right)$$
Very few patterns have been found in $OEIS$.