These are my left cosets so far, according to Lagrange's Theorem I am missing one more left coset right?
\begin{align*} \\(1234)H=&\lbrace (1234), (13), (1432), (24) \rbrace=(13)H=(1432)H=(24)H \\(12)H=&\lbrace (12), (34), (1324), (1423) \rbrace=(34)H=(1324)H=(1423)H \\(23)H=&\lbrace (23), (1342), (1243), (14) \rbrace=(1342)H=(1243)H=(14)H \\(132)H=&\lbrace (132), (234), (124), (143)\rbrace=(234)H=(124)H=(143)H \\(123)H=&\lbrace (123), (134), (243), (142) \rbrace=(134)H=(243)H=(142)H \end{align*}
Of course : you are missing $H$ itself, aren't you? In your own notation,$$ \epsilon H=H=(12)(34)H=(13)(24)H=(14)(23)H $$ completes a list of $24$ elements of $S_4$.
I should add that the way you enumerate the elements of every coset is very useful, because it gives you candidates which can generate remaining cosets. For example, when you wrote down five lines and $20$ elements, I had to only see which elements were missing from the list of elements to see all the remaining elements of the last coset. This turned out to be $H$ itself, which is always a left/right coset of itself, as you will not forget from now on!