Globally, I do not know of a global name, because it is either a median or a (locally) linear filter with fluctuating tap locations: a median (odd length), a 3-tap Gaussian approximation (2-length) or a weighted average of at most 4 samples.
- when the length is odd ($2l+1$): a simple median filter
- when the length is two: a simple Pascal averaging linear filter with taps $[1\,,2\, ,1]/4$
when the length is even ($2l$), greater than or equal to 4: a linear filter with up to 4 non-zero coefficients, but with different locations (depending on the order of the samples) across an $l+1$ window, with one of the following shapes:
- $[\cdots,1\,\cdots,1\,\cdots,1\,\cdots, 1,\cdots]/4$
- $[\cdots,1\,\cdots,2\,\cdots, 1,\cdots]/4$
- $[\cdots,1\,\cdots,1\,\cdots]/2$
I could not have your initial code work on my computer. Let me offer a more compact version.
data = Signal';
lFilter = 10;
dataDoubleMedianFilter = mean([medfilt1(data,length),flipud(medfilt1(flipud(data),length))],2);
Due to the symmetry in the median filter with odd length, your filter is just a simple median filter in that case. In other words, you average the two same median filtered signals.
If you set an even length, as you did, it is a little more involved. Morally, I do not understand the rationale. Even-length median "invents" novel values, stemming from the mean on the two centered samples.
Let us first try with a 2-median. On a centered set of values $[x_{-1}\,x_{0}\,x_{1}]$, one media will give $(x_{-1}+x_{0})/2$, the other $(x_{0}+x_{1})/2$, and their mean is the linear Pascal filter of length $3$, with taps $[1\,,2\,, 1]/4$:
$$(x_{-1}+2x_{0}+x_{1})/4$$
More generally, for an even number $2l$ of samples, the direct and the reverse frames share a center of $2l-1$ samples $x_1,\ldots,x_{2l-1}$, and differ from the 2 outer samples $x_0$ and $x_{2l}$. The resulting filter will yields the mean of the two most-centered samples from the left and the right frame. So it will be linear in aspect. However, the filter taps may have different locations. For instance, if the three-most centered samples (call them $x_{\sigma{(1)}},x_{\sigma{(2)}},x_{\sigma{(3)}}$ after reordering) of the $2l+1$ total frame belong to the $2l-1$ center frame, then they will be averaged with weights $1/4$, $1/2$, $1/4$:
$$1/4x_{\sigma{(1)}} + 1/2x_{\sigma{(2)}}+1/4x_{\sigma{(3)}}$$
If the outer samples reshuffle the order of the center frame, other combinations are obtained, with coefficients $1/2$, $1/2$ or $1/4$, $1/4$, $1/4$, $1/4$.