= Solution
Take the <Jacobi theta functions> at zero, with $q=e^{\pi i\tau}$:
$$
\theta_2(\tau)=\sum_{n\in\mathbb Z}q^{(n+1/2)^2},\quad
\theta_3(\tau)=\sum_{n\in\mathbb Z}q^{n^2},\quad
\theta_4(\tau)=\sum_{n\in\mathbb Z}(-1)^nq^{n^2}.
$$
Define the <modular lambda function> by
$$
\boxed{\lambda(\tau)=\frac{\theta_2(\tau)^4}{\theta_3(\tau)^4}.}
$$
This is an explicit holomorphic expression on the <complex upper half-plane>. The product forms
$$
\begin{aligned}
\theta_2&=2q^{1/4}\prod_{m\geq1}(1-q^{2m})(1+q^{2m})^2,\\
\theta_3&=\prod_{m\geq1}(1-q^{2m})(1+q^{2m-1})^2,\\
\theta_4&=\prod_{m\geq1}(1-q^{2m})(1-q^{2m-1})^2
\end{aligned}
$$
show that all three are nonzero there. The theta identity $\theta_3^4=\theta_2^4+\theta_4^4$ gives $1-\lambda=\theta_4^4/\theta_3^4$, so $\lambda$ omits $0,1$ and infinity on the half-plane.
The <theta-constant inversion and translation laws>, obtained from <Poisson summation> and changing indices in the defining sums, give
$$
\lambda(-1/\tau)=1-\lambda(\tau),\qquad
\lambda(\tau+1)=\frac{\lambda(\tau)}{\lambda(\tau)-1}.
$$
These formulas imply invariance under the projective group
$$
\overline{\Gamma(2)}
=\left\{\begin{pmatrix}a&b\\c&d\end{pmatrix}\in SL_2(\mathbb Z):
a,d\text{ odd},\ b,c\text{ even}\right\}/\{\pm I\},
$$
generated by $\tau\mapsto\tau+2$ and $\tau\mapsto\tau/(2\tau+1)$. This is the effective level-two <principal congruence subgroup>.
Its action is properly discontinuous because it is a discrete subgroup of $PSL_2(\mathbb R)$. It is also free. An elliptic integral matrix has trace $0$ or $\pm1$. The odd traces are excluded because $a+d$ is even. Trace zero would give $d=-a$ and $bc=-a^2-1\equiv2\pmod4$, impossible when $b,c$ are even. Thus the only elements acting trivially are the already quotiented central matrices $\pm I$.
A <fundamental domain> is the ideal quadrilateral
$$
-1<\operatorname{Re}\tau<1,\qquad
|\tau-1/2|>1/2,\qquad|\tau+1/2|>1/2,
$$
with vertices $-1,0,1,\infty$. The vertical sides are paired by translation by two, and the semicircular sides by $\tau\mapsto\tau/(2\tau+1)$. Reduction using these pairings gives the domain; equivalently it is a union of six translates of the standard modular fundamental domain, consistent with the six elements of $PSL_2(\mathbb F_2)$. On adjoining the cusp points, there are three vertex classes, two paired edges and one face. The resulting orientable compact surface has Euler characteristic $3-2+1=2$, hence is a sphere. This is the compactification of the <level-two modular curve>.
All three cusp widths are two, so their local parameters are obtained from $q=e^{\pi i\tau}$ after a modular change of coordinate. The defining products give
$$
\lambda(\tau)=16q-128q^2+O(q^3)
$$
at infinity: there is a simple zero. At the cusp zero, inversion gives $1-\lambda=16q_0+O(q_0^2)$, so the value is $1$. At the cusp one, combine inversion and translation; the value is infinity and the pole is simple. The three cusp classes are precisely the parity classes of primitive rational pairs $(p,q)$: odd-even, even-odd and odd-odd. A Bézout identity, with parity adjusted by adding multiples of $(p,q)$, places any such cusp in the orbit of $\infty$, $0$ or $1$, respectively.
The descended <meromorphic function> on the compactified quotient thus has exactly one simple pole. Its degree is one, since the degree equals its total pole multiplicity; it is therefore a <biholomorphic> coordinate on that sphere. Removing the three cusps identifies the quotient with $\mathbb C\setminus\{0,1\}$. Consequently
$$
\boxed{\lambda:\mathbb H\longrightarrow\widehat{\mathbb C}\setminus\{0,1,\infty\}}
$$
is a <covering map> with deck group $\overline{\Gamma(2)}$. The source half-plane is simply connected, so it is the required <universal covering map>. This proves the <universal covering by the modular lambda function>, including the absence of interior branching. Three arbitrary distinct punctures are handled by a target <Möbius transformation>.
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