I know that if $L$ is a linear transformation from $V$ to $W$ where $V,W$ are finite dimensional, then we can conclude that the dimension of image (rank) of $L$ is same as that of its transpose, i.e., $L^t$.
But what happens when: $\dim V,\dim W=\infty$, or just one of them has infinite dimension? If there is any difference in the above statement, why such difference arises?
There are two possible answers to your question.
First I will assume that you don't distinguish between different sizes of infinity.
Let $L \colon V \to W$ of infinite rank be given, and let $n$ be a natural number. It is possible to find a finite-dimensional subspace $V_1$ of $V$ and a finite dimensional quotient $W_1$ of $W$ such that the composite mapping $V_1 \to V \to W \to W_1$ has rank $n$. The transpose of this mapping is $W_1' \to W' \to V' \to V_1'$, and this composite has rank $n$ by the finite-dimensional theorem. But the middle arrow is $L^{t}$, and it must therefore have rank $\geq n$. Since $n$ was arbitrary, $L^{t}$ in fact has infinite rank.
Now we look at the question in a way that distinguishes between different sizes of infinity. Let $V = \mathbb{Q}^{(\mathbb{N})}$ be the sum of countably many copies of $\mathbb{Q}$, and let $L$ be the identity on $V$. The transpose of $L$ is the identity on $V' = \mathbb{Q}^{\mathbb{N}}$, which has uncountable dimension. Thus the rank of $L^{t}$ is strictly greater than that of $L$.