For any valid sequence $Y_1,\ldots,Y_7$ the two equations $ \sum_{i=1}^7Y_i=0 $ and $ \sum_{i=1}^7a^{i-1}Y_i=0 $ should hold. This is seen as follows. These sums compute the values of the encoded polynomial $Y(x)=Y_1+Y_2x+\cdots+Y_7x^6$ at the points $x=1$ and $x=a$ respectively. They are both zeros of $Y(x)$, because the factors $x-1=x+1$ and $x-a=x+a$ appear in the defining equation $ Y(x)=(x+1)(x+a)M(x). $ Here $M(x)=\sum_{i=1}^5X_ix^{i-1}$ is the message polynomial. We see that these equations do not hold for the received sequence $\tilde{Y}_i, i=1,\ldots,7$, because $S(1)=\sum_{i=1}^7\tilde{Y}_i=(1,0,1,1,1)$ is not zero. Therefore at least one of the symbols $\tilde{Y}_i,i=1,\ldots,7$ is in error. Assume that only a single error happened, so there exists an index $i_0$ such that $\tilde{Y}_{i_0}=Y_{i_0}+E$, where $E\neq0$, and $\tilde{Y}_i=Y_i$ for all $i\neq i_0$. Compute also $ S(a)=\sum_{i=1}^7a^{i-1}\tilde{Y}_i. $ I cannot do this for you, because you didn't tell me whether you use little-endian or big-endian notation. IOW I don't know, whether for example your $\tilde{Y}_2=(1,0,1,0,0)$ is equal to $a^0+a^2$ or $a^4+a^2$.
Once you have $S(1)$ and $S(a)$, then the rest is straightforward. Because these checksums are linear combinations of the $Y$s, and they vanish whenever we have a valid codeword, we have that $ S(1)=\sum_{i=1}^7\tilde{Y}_i=E+\sum_{i=1}^7Y_i=E+0=E, $ and $ S(a)=\sum_{i=1}^7a^{i-1}\tilde{Y}_i=a^{i_0-1}E+\sum_{i=1}^7a^{i-1}Y_i=a^{i_0-1}E. $ We want determine $i_0$, because that tells us which symbol $\tilde{Y}_i$ is corrupt. And we want to solve $E$, because that is the 'error amount'. From the preceding equations you can compute $ a^{i_0-1}=\frac{a^{i_0-1}E}{E}=\frac{S(a)}{S(1)}. $ Because you know both $S(1)$ and $S(a)$, this is easy to calculate. The answer should be one of the elements $1,a,a^2,\ldots,a^6$. Just compute these elements of $GF(32)$ and look for match. If you find one, then you know $i_0-1$, and hence the location of the errorneous coefficient. If there is no match, then more than one error has occured.
Of course, the error value $E$ is then equal to $S(1)$. So you simply add $E=S(1)$ to the coefficient $\tilde{Y}_{i_0}$, because $Y_{i_0}=\tilde{Y}_{i_0}+E$.
At the end you extract the polynomial $M(x)$ e.g. by long division as Dilip pointed out.