You want to solve the equation $(I+X)^2=I+Y$ for the $(n\times n)$-matrix $X$ when $Y$ is a given matrix near $0$. This amounts to
$Z(X):=2X+X^2\ =\ Y\ ,\qquad (*)$
an equation in the $n^2$ unknowns $X_{ik}$. When $Y=0$ then $X=0$ is a solution of this equation. According to the assumptions of the implicit function theorem we have to check whether the $(n^2\times n^2)$-matrix
$J:=\Bigl({\partial(Z_{11},Z_{12},\ldots,Z_{nn})\over \partial(X_{11},X_{12},\ldots,X_{nn})}\Bigl)_{X=0}$
is regular. As $Z_{ik}=2 X_{ik} +$ terms of degree $2$ in the $X_{ik}$, the matrix $J$ is $2$ times the $n^2\times n^2$ identity matrix; so it is certainly regular. It follows that the equation $(*)$ has a solution $X=F(Y)$ such that $F(0)=0$ and $F(\cdot)$ is a differentiable matrix valued function of (the matrix elements of) $Y$.
By the way, the square root $I+X$ of $I+Y$ for $\|Y\|<1$ is given by the binomial series
$\sum_{k=0}^\infty\ {1/2 \choose k}\ Y^k\ .$